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How to Solve Equations with Variables on Both Sides
Last Updated: March 11, 2023 Fact Checked
This article was co-authored by JohnK Wright V . JohnK Wright V is a Certified Math Teacher at Bridge Builder Academy in Plano, Texas. With over 20 years of teaching experience, he is a Texas SBEC Certified 8-12 Mathematics Teacher. He has taught in six different schools and has taught pre-algebra, algebra 1, geometry, algebra 2, pre-calculus, statistics, math reasoning, and math models with applications. He was a Mathematics Major at Southeastern Louisiana and he has a Bachelor of Science from The University of the State of New York (now Excelsior University) and a Master of Science in Computer Information Systems from Boston University. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 186,481 times.
To study algebra, you will see equations that have a variable on one side, but later on you will often see equations that have variables on both sides. The most important thing to remember when solving such equations is that whatever you do to one side of the equation, you must do to the other side. Using this rule, it is easy to move variables around so that you can isolate them and use basic operations to find their value.
Solving Equations with One Variable on Both Sides
Solving System Equations with Two Variables
Solving Example Problems
Community Q&A
Things You'll Need
You might also like.
- ↑ https://www.khanacademy.org/math/cc-sixth-grade-math/cc-6th-expressions-and-variables/cc-6th-distributive-property/v/the-distributive-property
- ↑ https://www.virtualnerd.com/algebra-1/linear-equations-solve/variables-both-sides-equations/variables-both-sides-solution/variables-grouping-symbols-both-sides
- ↑ https://www.youtube.com/watch?v=hrAOSknrYiI&t=296s
- ↑ https://www.khanacademy.org/math/cc-sixth-grade-math/cc-6th-expressions-and-variables/cc-6th-evaluating-expressions/v/expression-terms-factors-and-coefficients
- ↑ https://www.virtualnerd.com/pre-algebra/linear-functions-graphing/system-of-equations/solving-systems-equations/two-equations-two-variables-substitution
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2.4.7: Equations with Variables on Both Sides
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Equations with Variables on Both Sides
Bill and Kate are both reading a 500-page novel. So far, Bill has read 70 pages and Kate has read 50 pages, but from this point forward, Bill plans to read 25 pages per day, while Kate plans to read 29 pages per day. After how many days will they have read the same number of pages? Do you know how to set up and solve an equation to answer a question like this?
Solving Equations with Variables on Both Sides
As you may now notice, equations come in all sizes and styles. There are single-step, double-step, and multi-step equations. In this Concept, you will learn how to solve equations with a variable appearing on each side of the equation. The process you need to solve this type of equation is similar to solving a multi-step equation. The procedure is repeated here.
Procedure to Solve Equations:
Step 1: Remove any parentheses by using the Distributive Property or the Multiplication Property of Equality.
Step 2: Simplify each side of the equation by combining like terms.
Step 3: Isolate the ax term. Use the Addition Property of Equality to get the variable on one side of the equal sign and the numerical values on the other.
Step 4: Isolate the variable. Use the Multiplication Property of Equality to get the variable alone on one side of the equation.
Step 5: Check your solution.
Let's solve the following problems:
- Karen and Sarah have bank accounts. Karen has a starting balance of $125.00 and is depositing $20 each week. Sarah has a starting balance of $43 and is depositing $37 each week. When will the girls have the same amount of money?
To solve this problem, you could use the “guess and check” method. You are looking for a particular week in which the bank accounts are equal. This could take a long time! You could also translate the sentence into an equation. The number of weeks is unknown so this is our variable. Call it w. Now translate this situation into an algebraic equation:
125+20w=43+37w
This is a situation in which the variable w appears on both sides of the equation.
Using the Addition Property of Equality, move the variables to one side of the equation:
125+20w−20w=43+37w−20w
Simplify: 125=43+17w
Solve using the steps above:
125−43=43−43+17w
82÷17=17w÷17
w≈4.82
Checking your solution:
125+20(4.82)=43+37(4.82)
125+96.4=43+178.34
221.4≈221.34
Note that both sides of the equation are not exactly the same because we used an approximation to the check the solution. However, they are close enough and it will take about 4.8 weeks for Sarah and Karen to have equal amounts of money.
- Solve 2y−5=3y+10.
To solve this equation, we need to get both terms with a variable onto the same side. The easiest way to do this is to subtract the variable with the smaller coefficient from each side:
2y−5=3y+10
−2y+2y−5=−2y+3y+10
−5=y+10
−5−10=y+10−10
−15=y
2(−15)−5=3(−15)+10
−30−5=−45+10
−35=−35
- Solve for h:3(h+1)=11h−23.
First you must remove the parentheses by using the Distributive Property:
3h+3=11h−23
Gather the variables on one side:
3h−3h+3=11h−3h−23
3=8h−23
Solve using the steps from above:
3+23=8h−23+23
26÷8=8h÷8
h=13/4=3.25
3(3.25)+3=11(3.25)−23
9.75+3=35.75−23
12.75=12.75
Example 2.4.7.1
Earlier, you were told that Bill has read 70 pages and plans to read 25 pages per day while Kate has read 50 pages but plans to read 29 pages per day. How many days will it take for them to have read the same number of pages?
Let's represent the unknown number of days as the variable d.
The equation that represents this situation is:
70+25d=50+29d
where the left side of the equation represents the pages that Bill has read and the right side of the equation represents how many pages Kate has read.
First, move the variables to one side:
70+25d−25d=50+29d−25d
Now simplify using the steps from the beginning of this concept:
70−50=50+4d−50
20÷4=4d÷4
5=d Checking your solution:
70+25(5)=50+29(5)
70+125=50+145
It will take Bill and Kate 5 days until they have read the same amount of pages.
Example 2.4.7.2
Solve for g when −5g+3=−8g+9.
In this case, our two terms with the variable are negative. Since −8<−5, we will subtract -8. However, subtracting -8 can be simplified to adding 8 (Opposite-Opposite Property). This will leave us with a positive coefficient in front of our variable.
−5g+3=−8g+9+8g−5g+3=+8g−8g+9
3g+3−3=9−3
(1/3)⋅3g=(1/3)⋅6
Checking the solution:
−5g+3=−8g+9
−5(2)+3=−8(2)+9
−10+3=−16+9
−7=−7
Therefore, g=2.
In 1 – 13, solve the equation.
- 3(x−1)=2(x+3)
- 7(x+20)=x+5
- 9(x−2)=3x+3
- 2(a−(1/3))=(2/5)(a+(2/3))
- (2/7)(t+(2/3))=(1/5)(t−(2/3))
- (1/7)(v+(1/4))=2((3v/2)−(5/2))
- (y−4)/11=(2/5)⋅(2y+1)/3
- (z/16)=(2(3z+1))/9
- (q/16)+(q/6)=((3q+1)/9)+(3/2)
- 21+3b=6−6(1−4b)
- −2x+8=8(1−4x)
- 3(−5v−4)=−6v−39
- −5(5k+7)=25+5k
- Manoj and Tamar are arguing about how a number trick they heard goes. Tamar tells Andrew to think of a number, multiply it by five, and subtract three from the result. Then Manoj tells Andrew to think of a number, add five, and multiply the result by three. Andrew says that whichever way he does the trick he gets the same answer. What was Andrew's number?
- I have enough money to buy five regular priced CDs and have $6 left over. However, all CDs are on sale today for $4 less than usual. If I borrow $2, I can afford nine of them. How much are CDs on sale for today?
- Jaime has a bank account with a balance of $412 and is saving $18 each week. George has a bank account with a balance of $874 and is spending $44 dollars each week. When will the two have the same amount of money?
- At how many texts will the two plans charge the same?
- Suppose you plan to text 3,000 times per month. Which plan should you choose? Why?
- When will the two companies charge the same?
- You will need the tank for a 24-hour fund raise-a-thon. Which company should you choose?
Mixed Review
- Solve for t: −12+t=−20.
- Solve for r: 3r−7r=32.
- Solve for e: 35=5(e+2).
- 25 more than four times a number is 13. What is the number?
- Find the opposite of 9(1/5). Write your answer as an improper fraction.
- Evaluate (|b|−a)−(|d|−a). Let a=4, b=−6, and d=5.
- Give an example of an integer that is not a counting number.
- Determine the inverse of addition.
- Solve for w: −4w=16.
- Write an equation to represent the following situation and solve. Shauna ran the 400 meter dash in 56.7 seconds, 0.98 seconds less than her previous time. What was her previous time?
- Solve for b: (1/2)b+5=9.
- Solve for q: 3q+5−4q=19.
Review (Answers)
To see the Review answers, open this PDF file and look for section 3.6.
Additional Resources
PLIX: Play, Learn, Interact, eXplore: Rubber Ducky Math
Video: Identifying Solutions Given Variables on Both Sides - Overview
Activities: Equations with Variables on Both Sides Discussion Questions
Practice: Equations with Variables on Both Sides
Real World Application: The Cost of College
Module 1: Solving Linear Equations with One Variable
1.4 solving linear equations with variables on both sides of the equation, section 1.4 learning objectives.
1.4: Solving Linear Equations with Variables on Both Sides of the Equation
Solving linear equations with variables on both sides of the equation
Use both the distributive property and combining like terms to simplify and then solve algebraic equations with variables on both sides of the equation.
- Classifying solutions to linear equations
Some equations may have the variable on both sides of the equal sign, as in this equation: [latex]4x-6=2x+10[/latex].
To solve this equation, we need to “move” one of the variable terms. This can make it difficult to decide which side to work with. It doesn’t matter which term gets moved, [latex]4x[/latex] or [latex]2x[/latex], however, to avoid negative coefficients, you can move the smaller term.
Solve: [latex]4x-6=2x+10[/latex]
Choose the variable term to move—to avoid negative terms choose [latex]2x[/latex]
[latex]\begin{array}{r}4x-6=2x+10\\\underline{-2x\,\,\,\,\,\,\,\,\,\,-2x}\\\,\,\,2x-6=10\end{array}[/latex]
Now add 6 to both sides to isolate the term with the variable.
[latex]\begin{array}{r}2x-6=10\\\underline{\,\,\,\,+6\,\,\,+6}\\2x=16\end{array}[/latex]
Now divide each side by 2 to isolate the variable x.
[latex]\begin{array}{c}\frac{2x}{2}=\frac{16}{2}\\\\x=8\end{array}[/latex]
Recall the Distributive Property from last section
the distributive property of multiplication.
For all real numbers a, b, and c , [latex]a(b+c)=ab+ac[/latex].
What this means is that when a number multiplies an expression inside parentheses, you can distribute the multiplication to each term of the expression individually. Then, you can follow the steps we have already practiced to isolate the variable and solve the equation.
In the next example, you will see that there are parentheses on both sides of the equal sign, so you will need to use the distributive property (that we learned last section) twice. Notice that you are going to need to distribute a negative number, so be careful with negative signs!
Solve for [latex]t[/latex].
[latex]2\left(4t-5\right)=-3\left(2t+1\right)[/latex]
Apply the distributive property to expand [latex]2\left(4t-5\right)[/latex] to [latex]8t-10[/latex] and [latex]-3\left(2t+1\right)[/latex] to[latex]-6t-3[/latex]. Be careful in this step—you are distributing a negative number, so keep track of the sign of each number after you multiply.
[latex]\begin{array}{r}2\left(4t-5\right)=-3\left(2t+1\right)\,\,\,\,\,\, \\ 8t-10=-6t-3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]
Add [latex]6t[/latex] to both sides to begin combining like terms.
[latex]\begin{array}{r}8t-10=-6t-3\\ \underline{+6t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+6t}\,\,\,\,\,\,\,\,\,\\ 14t-10=\,\,\,\,-3\,\,\,\,\,\,\,\end{array}[/latex]
Add 10 to both sides of the equation to isolate t .
[latex]\begin{array}{r}14t-10=-3\\ \underline{+10\,\,\,+10}\\ 14t=\,\,\,7\,\end{array}[/latex]
The last step is to divide both sides by 14 to completely isolate t .
[latex]\begin{array}{r}14t=7\,\,\,\,\\\frac{14t}{14}=\frac{7}{14}\end{array}[/latex]
[latex]t=\frac{1}{2}[/latex]
We simplified the fraction [latex]\frac{7}{14}[/latex] into [latex]\frac{1}{2}[/latex]
In the following video, we solve another multi-step equation with two sets of parentheses.
In the next example, we will use both the Distributive Property and combining like terms to simplify before solving the equation.
Solve for [latex]x[/latex].
[latex]4\left(x-3\right)+2=2x-\left(x+1\right)[/latex]
In simplifying the right side of the equation, it may be helpful to think of [latex]-(x+1)[/latex] as [latex]-1(x+1)[/latex]
[latex]4\left(x-3\right)+2=2x-1\left(x+1\right)[/latex]
Use the distributive property to get rid of parentheses
[latex]4x-12+2=2x-x-1[/latex]
Combine like terms on each side of the equation
[latex]4x-10=x-1[/latex]
Subtract [latex]x[/latex] from both sides of the equation
[latex]\begin{array}{r}4x-10=x-1\\ \underline{-x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-x}\,\,\,\,\,\,\,\\ 3x-10=\,\,\,\,-1\,\,\,\,\,\,\,\end{array}[/latex]
Add 10 to both sides of the equation to isolate the x term
[latex]\begin{array}{r}3x-10=-1\\ \underline{+10\,\,\,+10}\\ 3x=\,\,\,9\,\end{array}[/latex]
The last step is to divide both sides by 3 to completely isolate x.
[latex]\begin{array}{r}3x=9\,\,\,\,\\\frac{3x}{3}=\frac{9}{3}\end{array}[/latex]
[latex]x=3[/latex]
Classifying solutions to Linear Equations
There are three cases that can come up as we are solving linear equations. We have already seen one, where an equation has one solution. Sometimes we come across equations that do not have any solutions and even some that have an infinite number of solutions. The case where an equation has no solution is illustrated in the next example.
Equations with No Solutions
Solve for [latex]x[/latex].
[latex]12+2x–8=7x+5–5x[/latex]
Combine like terms on both sides of the equation.
[latex] \displaystyle \begin{array}{l}12+2x-8=7x+5-5x\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,2x+4=2x+5\end{array}[/latex]
Isolate the x term by subtracting 2 x from both sides.
[latex]\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,2x+4=2x+5\\\,\,\,\,\,\,\,\,\underline{-2x\,\,\,\,\,\,\,\,\,\,-2x\,\,\,\,\,\,\,\,}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4= \,5\end{array}[/latex]
This false statement implies there are no solutions to this equation. Sometimes, we say the solution does not exist, or DNE for short.
Notice in the example above you did not find a value for x . Solving for x the way you know how, you arrive at the false statement [latex]4=5[/latex]. Surely [latex]4[/latex] cannot be equal to [latex]5[/latex]!
This may make sense when you consider the second line in the solution where like terms were combined. If you multiply a number by [latex]2[/latex] and add [latex]4[/latex] you would never get the same answer as when you multiply that same number by [latex]2[/latex] and add [latex]5[/latex]. Since there is no value of x that will ever make this a true statement, the solution to the equation above is “no solution.”
Be careful that you do not confuse the solution [latex]x=0[/latex] with “no solution.” The solution [latex]x=0[/latex] means that the value [latex]0[/latex] satisfies the equation, so there is a solution. “No solution” means that there is no value, not even [latex]0[/latex], which would satisfy the equation.
Also, be careful not to make the mistake of thinking that the equation [latex]4=5[/latex] means that [latex]4[/latex] and [latex]5[/latex] are values for x that are solutions. If you substitute these values into the original equation, you’ll see that they do not satisfy the equation. This is because there is truly no solution —there are no values for x that will make the equation [latex]12+2x–8=7x+5–5x[/latex] true.
Algebraic Equations with an Infinite Number of Solutions
You have seen that if an equation has no solution, you end up with a false statement instead of a value for x . It is possible to have an equation where any value for x will provide a solution to the equation. In the example below, notice how combining the terms [latex]5x[/latex] and [latex]-4x[/latex] on the left leaves us with an equation with exactly the same terms on both sides of the equal sign.
[latex]5x+3–4x=3+x[/latex]
[latex] \displaystyle \begin{array}{r}5x+3-4x=3+x\\x+3=3+x\end{array}[/latex]
Isolate the x term by subtracting x from both sides.
[latex]\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,x+3=3+x\\\,\,\,\,\,\,\,\,\underline{\,-x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-x\,}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\,\,=\,\,3\end{array}[/latex]
This true statement implies there are an infinite number of solutions to this equation, or we can also write the solution as “All Real Numbers”
In that last example, we arrived at the true statement “[latex]3=3[/latex].” When you end up with a true statement like this, it means that the solution to the equation is “all real numbers.” Try substituting [latex]x=0[/latex] into the original equation—you will get a true statement! Try [latex]x=-\dfrac{3}{4}[/latex] and it will also check!
This equation happens to have an infinite number of solutions. Any value for x that you can think of will make this equation true. When you think about the context of the problem, this makes sense—the equation [latex]x+3=3+x[/latex] means “some number plus [latex]3[/latex] is equal to [latex]3[/latex] plus that same number.” We know that this is always true—it is the commutative property of addition!
[latex]3\left(2x-5\right)=6x-15[/latex]
Distribute the [latex]3[/latex] through the parentheses on the left-hand side.
[latex] \begin{array}{r}3\left(2x-5\right)=6x-15\\6x-15=6x-15\end{array}[/latex]
Wait! This looks just like the previous example. You have the same expression on both sides of an equal sign. No matter what number you choose for x , you will have a true statement. We can finish the algebra:
[latex]\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,6x-15=6x-15\\\,\,\,\,\,\,\,\,\underline{\,-6x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-6x\,}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-15\,\,=\,\,-15\end{array}[/latex]
This true statement implies there are an infinite number of solutions to this equation.
In the following video, we show more examples of attempting to solve a linear equation with either no solution or many solutions.
In the following video, we show more examples of solving linear equations with parentheses that have either no solution or many solutions.
Home / United States / Math Classes / Solving Equations with Variables on Boths Sides
Solving Equations with Variables on Boths Sides
Some equations have variables on both sides of the equation. In such cases, the equations have different types of soluti ons. Learn how to use the properties of the equations to determine the nature of the solution of the equation and to find the solution. ...Read More Read Less
About Equations of Variables on Both Sides
Solving equations with variables on both sides of the equation
- Two sides of an equation
Solution of an equation
Solving an equation using properties of equalities, using the distributive property to solve an equation.
- Equations with no solution
Equations with infinitely many solutions
- Frequently Asked Questions
An equation is a mathematical statement that connects two equal expressions with an “equal to” sign. 2x + 5 = 25 – 2x, y – 6 = 10 – 3y, and 8z – 6 = 29 – z are some examples of equations having variables on both sides. The unknown values like x, y, and z found in these equations are known as variables. We can find the values of variables of an equation by rearranging the terms using some special properties.
Two Sides of an Equation
Every equation has two sides: the expression on the left-hand side (LHS) of the equation and the expression on the right-hand side of the equation (RHS). In every true equation, the left-hand side will be equal to the right-hand side (LHS = RHS).
A solution of an equation is a value that can be used in the place of the variable to make a true statement. For example, the solution of x in the equation x – 2 = 0 is 2. If we use 2 in the place of x , we get (2) – 2 = 0 , which is a true statement. Some equations have only one solution. Certain equations have no solution. And certain equations have infinitely many solutions.
While solving an equation, we are essentially finding the value of the unknown variable. So, the solution of an equation is the value of the variable. All equations can be solved by simplifying the terms on both sides and isolating the variable to find its solution. We need to use inverse math operations to isolate the variable. The general procedure for solving multi-step equations is as follows:
First, we use the distributive property to expand the terms inside the parentheses. Then, we combine the like terms and simplify the equation. Finally, we isolate the variable using inverse operations.
For example , let’s solve the equation y – 6 = 10 – 3y .
y – 6 = 10 – 3y Write the equation
Add 6 on both sides using the addition property of equality to simplify the equation.
y – 6 + 6 = 10 – 3y + 6 Subtraction property of equality
y = 16 – 3y Subtract
Add 3y on both sides using the addition property of equality to simplify the equation.
y + 3y = 16 – 3y + 3y Subtraction property of equality
4y = 16 Subtract
Divide both sides by 4 using the division property of equality.
\(\frac{4y}{4}\) = \(\frac{16}{4}\) Division property of equality
y = 4 Simplify
So, the value of y in this equation is 4.
The distributive property states that multiplying the sum of two or more addends by a number is the same as multiplying each addend individually by the number and then adding the products together. We can use distributive property to expand the terms inside the parentheses.
Suppose you want to solve the equation 2 ( 2x + 5 ) = 25 – x . The first step is to expand the terms inside the parentheses using distributive property.
2 ( 2x + 5 ) = 25 -x Write the equation
4x + 10 = 25 – x Distributive property
Now, we combine the like terms together. We can add x on both sides.
4x + 10 + x = 25 -x + x
5x + 10 = 25 Add
We can subtract 10 on both sides using the subtraction property of equality.
5x + 10 – 10 = 25 – 10
5x = 15 Subtract
Now, divide both sides by 5 using the division property of equality to isolate the variable.
\(\frac{5x}{5}\) = \(\frac{15}{5}\) Division property of equality
x = 3 Simplify
So, the solution of this equation is x = 3 .
Equations with no Solution
Unlike the examples we solved before, certain equations may not have any solution. When solving such equations, we will get an equivalent equation that is not a true statement for any value of the variable. For example, the final equation or statement might be 5 = 0 , which is not true.
5x + 3 = 5x + 5 is an example of an equation that has no solution.
The first step in solving the equation is to group the like terms.
5x + 3 = 5x + 5 Write the equation
5x + 3 – 5x = 5x + 5 – 5x Subtraction property of equality
We write the final equation as 3 = 5 , which is not a true statement. Hence, the equation 5x + 3 = 5x + 5 does not have any solution.
When solving equations with infinitely many solutions, we will get an equivalent solution that is true for all values of the variable. 8x + 3 = 8x + 3 is an example of an equation with infinitely many solutions. To solve this equation, we can group the common terms.
8x + 3 = 8x + 3 Write the equation
-8x -8x Subtraction property of equality
The solution 3 = 3 is true for all values of the variable. Hence, the equation has infinitely many solutions.
Example 1: Solve 5 ( x – 1 ) = -4
5 ( x – 1 ) = -4 Write the equation
5x – 5 = -4 Distributive property
5x – 5 + 5 = -4 + 5 Addition property of equality
5x = 1 Add
\(\frac{5x}{5}\) = \(\frac{1}{5}\) Division property of equality
\(x=\frac{1}{5}\) Simplify
Therefore, the solution of the equation is \(x=\frac{1}{5}\) .
Example 2: Solve 4 ( x – 3 ) = 4x – 12 .
4 ( x – 3 ) = 4x – 12 Write the equation
4x – 12 = 4x – 12 Distributive property
4x – 12 + 12 = 4x – 12 + 12 Addition property of equality
4x = 4x Add
The equation has an infinite number of solutions as it is true for all values of x .
Example 3: Solve 8x – 3 = 8x + 5 .
8x – 3 = 8x + 5 Write the equation
8x – 3 – 8x = 8x + 5 – 8x Subtraction property of equality
-3 = 5 Subtract
The final equation is not a true statement. Hence, the equation has no solution.
Example 4: A train starts its journey from New York with 250 passengers, stops at Philadelphia, and then reaches its destination at Washington. The number of passengers who boarded the train at Philadelphia is 3 times the number of passengers who got down at Philadelphia. Find the number of passengers who boarded the train from Philadelphia if 320 passengers got down at the final destination.
Number of passengers at the beginning of the journey = 250
Let the number of passengers who got down at Philadelphia be x .
So, the number of passengers who boarded from Philadelphia is 3x .
Number of passengers who got down at the final destination = 320
An equation relating the number of passengers can be formed as follows:
250 – x + 3x = 320
Here, we are subtracting x from 250 because x number of passengers got down at Philadelphia. At the same time, 3x passengers boarded the train from the same station. So, we add 3x to the left hand side of the equation.
250 – x + 3x = 320 Write the equation
250 + 2x = 320 Group the like terms
250 + 2x – 250 = 320 – 250 Subtraction property of equality
2x = 70 Subtract
\(\frac{2x}{2}\) = \(\frac{70}{2}\) Division property of equality
x = 35 Simplify
The solution of the equation 250 – x + 3x = 320 is 35.
So, 35 passengers got down at Philadelphia, and 3 x 35 = 105 passengers boarded the train from the same station.
What is the difference between an equation and an expression?
An expression is a mathematical phrase that contains numbers and variables. Expressions will not have an “equal to” sign. An equation is a mathematical statement that connects two equal expressions with an “equal to” sign
Can a linear equation have 3 solutions?
No, a linear equation has either a unique solution, no solution, or infinitely many solutions.
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Solving Equations having the Variable on Both Sides
How to Solve a Linear Equation Having Variables on Both Sides?
An equation consists of two expressions separated by an equal sign “=”. If an expression is represented using an equal sign, the value of one side should be equal to the value on the other side.
Now, previously, we only solved equations having one variable like $2x + 1 = 3$. This equation has a variable on one side as you can see only 2 has the variable x. But, now, we will solve equations having the variable on both sides. For illustration, we take this equation $4x + 2 = 3 - 9x$. In this equation, we have variables on both sides.
Solving equations having the variable on both sides means that there are two same variables in the equation: one is on the left side of the equation and other is on the right side of the equation; for illustration, we take this equation $3x + 2 = 3 - 2x$. In this equation, we have variables on both sides with some constant . These equations can be solved with many methods like addition and subtraction . Lets’s understand the topic by considering the following example.
Solve the given equation: $5x + 2 = 12$
$x = \dfrac{{10}}{5}$
Here, by solving the following equation, we get the value of the variable i.e. $x=2$.
The second example for variables on both sides:
Solve the given equation: $7x + 49 = 14x$
$7x+49=14x$
$49=14x-7x$
$x = \dfrac{{49}}{7}$
Here, by solving the following equation, we get the value of the variable i.e. $x=7$.
Solve for x Equations
Solve for x is meant to find the value of x in an equation of one variable that is x. Suppose we have an equation $5x - 10 = 0$ and are asked to determine the value of the variable. If the given variable is x in the equation, it is called solve for x. When we find the value of x and substitute it in the equation, we should get L.H.S = R.H.S.
Now illustrate this with some examples:
Solve for \[x:\dfrac{2}{5} = \dfrac{x}{{10}}\]
Cross multiply the fractions ,
\[ \Rightarrow 2 \times 10 = x \times 5\]
Solve the equation for x,
\[\Rightarrow x = \dfrac{{20}}{5}\]
Simplify for x,
\[ \Rightarrow x = 4\]
To verify the x value, put the result 4 back into the given equation,
\[\Rightarrow \dfrac{2}{5} = \dfrac{4}{{10}}\]
Cross multiply the fractions,
\[ \Rightarrow 2 \times 10 = 4 \times 5\]
\[\Rightarrow 20=20\]
L.H.S = R.H.S
Solve for Variables
Now we will learn how to solve an equation for a variable. Its steps are similar to solving for x. In solving for x, we only find the value of x but in solving for a variable, we have to find the value of every variable given in the equations. It generally has two equations having two variables in each equation.
Let’s take a look at an example:
Solve for the variables “x” and “y“: $2x - y = 5, 3x + 2y = 11$
\[ \Rightarrow 2x-y = 5\]
Adding y on both sides, we get,
\[ \Rightarrow 2x-y+y = 5+y\]
\[ \Rightarrow 2x = 5+y\]
\[ \Rightarrow x = \dfrac{({5+y)}}{2}\]
The above equation is known as x in terms of y.
Substitute \[x = \dfrac{{(5+y)}}{2}\] in the second equation \[x = \dfrac{{3(5+y)}}{2}+2y=11\]
\[\Rightarrow \dfrac{{3(5+y)}}{2}+2y=11\]
\[\Rightarrow \dfrac{{(15+3y+4y)}}{2}=11\]
\[\Rightarrow \dfrac{{(15+7y)}}{2}=11\]
\[\Rightarrow 15+7y=22\]
\[\Rightarrow 7y=22-15\]
\[\Rightarrow 7y=7\]
\[\Rightarrow y=1\]
Now, substitute y = 1 in \[x = \dfrac{({5+y)}}{2}\]
\[ \Rightarrow x = \dfrac{({6)}}{2}=3\]
Here, we get the variables of the given system of equations as x = 3 and y = 1.
Solving Equations having Variables on Both Sides
Consider the equation $5x – 4 = 2x + 2$. To evaluate the variable, we need to get all the variable terms to one side and the constant terms to the other side. Next, we combine like terms and then evaluate the variable by multiplying or dividing the expressions.
Step 1: Add and subtract terms to get the variables on one side and the constants on the other.
Step 2: Multiply or divide to evaluate the variable.
Example: Solve $5x – 4 = 2x + 2$.
Step 1: Get all the variable terms to one side and the constant terms to the other side.
$5x – 4 = 2x + 2$
$5x – 4 – 2x + 4 = 2x + 2 – 2x + 4$ (Subtract 2x and add 4 to both sides)
Step 2: Combine like terms.
$5x – 2x = 2 + 4$
Step 3: Divide or multiply to isolate the variable.
$3x = 6$ (Divide by 3)
Solving an Equation
If an equation is satisfied after the variable has been replaced by the solution or the value of variables, then the number is called a solution of the equation. The real number 3 is a solution of the equation $2x-1 = x+2$ since (2)(3) - 1=3+2. Also, 1 is the solution of the equation $(x-1)(x+2) = 0$ . A Solution is a value we put in place of a variable that makes the equation true.
For example: Find the solution of the given equation $2x-1 = x+2$.
\[ \Rightarrow 2x-1 = x+2\]
Add $(1-x)$ both sides of the equation.
\[ \Rightarrow 2x-1+1-x = x+2+1-x\]
\[ \Rightarrow x = 2+1\]
\[ \Rightarrow x = 3\]
The solution to the equation is $x=3$
Interesting Facts
Some interesting facts about equations having the variable on both sides are pointed out below.
If the coefficients of x are the same on both sides of the equation, then the sides will not equal each other; therefore, no solutions will occur.
Two equations having the same solution are called equivalent equations.
An equation having a single type of variable is called a linear equation.
Sir William Rowan Hamilton first discovered the linear equation.
Solved Problems
1) S olve for x in the given equation \[\dfrac{2-3x}{8} =\dfrac{8-x}{12}-\dfrac{11}{12}\].
\[\Rightarrow \dfrac{2-3x}{8} =\dfrac{8-x}{12}-\dfrac{11}{12}\]
\[\Rightarrow \dfrac{2-3x}{8} =\dfrac{8-x-11}{12}\]
\[\Rightarrow \dfrac{2-3x}{8} =\dfrac{x-3}{12}\]
Cross multiplying the expressions, we get,
\[\Rightarrow (2-3x)(12) =(x-3)(8)\]
\[\Rightarrow (24-36x) =(8x-24)\]
Adding $(24+36x)$ on both sides, we get,
\[\Rightarrow (24-36x+24-36x) =(8x-24+24+36x)\]
\[\Rightarrow (48) =(44x)\]
\[\Rightarrow \dfrac{48}{44} =x\]
\[\Rightarrow \dfrac{12}{11} =x\]
2) The sum of the two numbers is 25. One of the numbers exceeds the other by 9. Find the numbers.
Let the number be x.
Then, the other number became = 9+x
Now, the sum of the two numbers is = 25
According to question, x + x + 9 = 25
\[\Rightarrow 2x+9=25\]
\[\Rightarrow 2x=25-9\]
\[\Rightarrow 2x=16\]
\[\Rightarrow \dfrac{2x}{2}=\dfrac{16}{2}\] (divide by 2 into both sides)
\[\Rightarrow x=8\]
Therefore, the First number is 8.
The second number becomes = x + 9 = 8 + 9 = 17.
Therefore, the two numbers are 8 and 17.
Key Features
A linear equation has only one or two variables.
No variable in a linear equation has a power greater than 1.
We can calculate the value of the variable from the linear equation.
If a linear equation consists of two variables, then we can write a variable in terms of another variable.
Practice Questions
1. Solve for x : 3(2x – 4) = 4(2x + 4)
2. A positive number is 5 times another number. If 21 is added to both numbers, then one of the new numbers becomes twice the other new numbers. Find the original numbers.
Ans: 7 and 35
3. If the value of (3 + 2x) is equal to (3 - 2x), then what is the value of 5 + 3x?
FAQs on Solving Equations having the Variable on Both Sides
1. What is the degree of a linear equation?
The highest power of the variable is known as the degree of the equation. The highest power of the variable for a linear equation is 1. Thus, the degree of a linear equation is 1.
2. What is the number of solutions of a linear equation having two variables?
The linear equation having two variables is ax + by + c =0 where a and b are any real number and are not equal to zero. The solutions of y are \[y =- \dfrac{ax + c}{b}\]. Since x is a variable, the value of x is any real number. For each value of x, we get a unique value of y. The number of real numbers is infinity. Thus, the number of solutions of a linear equation having two variables is infinity.
3. What is the difference between linear equations and polynomials?
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Solving Linear Equations with Variables on Both Sides: A Step-by-Step Guide
Solving linear equations with variables on both sides is a fundamental skill in algebra that I consider essential for understanding a wide range of mathematical and real-world problems.
An equation of this sort typically looks like $ax + b = cx + d$, where $x$ i s the variable you’re looking to solve for, and $a$, $b$, $c$ , and $d$ represent constants.
My general strategy for tackling these equations involves a series of steps to isolate the variable on one side of the equation.
This means I’ll be adding, subtracting, multiplying, or dividing throughout to get $x$ by itself, which simplifies the equation down to something like $x = e$ , with $e$ being the solution.
In dealing with these types of equations, it’s important to maintain the balance of the equation , as whatever I do to one side, I must also do to the other to preserve equality.
Sometimes, this involves combining like terms or using the distributive property to simplify each side of the equation ; it’s like a mathematical dance where precision and balance are key.
Remember to check your solution by substituting it back into the original equation to ensure both sides equal out. Stay tuned as I walk you through this interesting and engaging process, sparking a light of understanding in the elegant dance of algebra.
Steps for Solving Equations With Variables on Both Sides
When I encounter an algebraic equation with variables on both sides, my first step is to simplify each side of the equation if needed. This involves expanding expressions using the distributive property , combining like terms , and organizing the equation so it becomes easier to work with.
Here’s a breakdown in a table format:
Let’s consider the equation, $7y = 11x + 4x – 8$. I would combine the $x$ terms on the right first, getting $7y = 15x – 8$. Now, if I need $y$ on one side and $x$ on the other, I might decide to move the $x$ terms to the other side by subtracting $15x$ from both sides, resulting in $7y – 15x = -8$.
After that, if there are any coefficients attached to the variable I’m solving for, I’ll divide both sides by that coefficient. If I am solving for $y$, and the equation is $7y – 15x = -8$, I’d divide everything by 7 to isolate $y$:
$$ \frac{7y}{7} – \frac{15x}{7} = \frac{-8}{7} $$ $$ y – \frac{15x}{7} = \frac{-8}{7} $$
Now, if I need to solve for $y$ explicitly, I might rearrange the terms to show $y$ as a function of $x$, which would give me $y = \frac{15x}{7} – \frac{8}{7}$ as the final step.
This straightforward approach ensures the equation is balanced and leaves me with a clear solution for the variable in question.
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9.3: Solve Equations with Variables and Constants on Both Sides
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Learning Objectives
By the end of this section, you will be able to:
- Solve an equation with constants on both sides
- Solve an equation with variables on both sides
- Solve an equation with variables and constants on both sides
Before you get started, take this readiness quiz.
- Simplify: 4y−9+9. If you missed this problem, review Exercise 1.10.20 .
Solve Equations with Constants on Both Sides
In all the equations we have solved so far, all the variable terms were on only one side of the equation with the constants on the other side. This does not happen all the time—so now we will learn to solve equations in which the variable terms, or constant terms, or both are on both sides of the equation.
Our strategy will involve choosing one side of the equation to be the “variable side”, and the other side of the equation to be the “constant side.” Then, we will use the Subtraction and Addition Properties of Equality to get all the variable terms together on one side of the equation and the constant terms together on the other side.
By doing this, we will transform the equation that began with variables and constants on both sides into the form \(ax=b\). We already know how to solve equations of this form by using the Division or Multiplication Properties of Equality.
Example \(\PageIndex{1}\)
Solve: \(7x+8=−13\).
In this equation, the variable is found only on the left side. It makes sense to call the left side the “variable” side. Therefore, the right side will be the “constant” side. We will write the labels above the equation to help us remember what goes where.
Since the left side is the “xx”, or variable side, the 8 is out of place. We must “undo” adding 8 by subtracting 8, and to keep the equality we must subtract 8 from both sides.
Try It \(\PageIndex{2}\)
Solve: \(3x+4=−8\).
\(x=−4\)
Try It \(\PageIndex{3}\)
Solve: \(5a+3=−37\).
\(a=−8\)
Example \(\PageIndex{4}\)
Solve: \(8y−9=31\).
Notice, the variable is only on the left side of the equation, so we will call this side the “variable” side, and the right side will be the “constant” side. Since the left side is the “variable” side, the 9 is out of place. It is subtracted from the 8y, so to “undo” subtraction, add 9 to both sides. Remember, whatever you do to the left, you must do to the right.
Try It \(\PageIndex{5}\)
Solve: \(5y−9=16\).
Try It \(\PageIndex{6}\)
Solve: \(3m−8=19\).
Solve Equations with Variables and Constants on Both Sides
The next example will be the first to have variables and constants on both sides of the equation. It may take several steps to solve this equation, so we need a clear and organized strategy.
Example \(\PageIndex{7}\)
Solve: \(9x=8x−6\).
Here the variable is on both sides, but the constants only appear on the right side, so let’s make the right side the “constant” side. Then the left side will be the “variable” side.
Try It \(\PageIndex{8}\)
Solve: \(6n=5n−10\).
\(n = -10\)
Try It \(\PageIndex{9}\)
Solve: \(-6c = -7c - 1\)
Example \(\PageIndex{10}\)
Solve: \(5y - 9 = 8y\)
The only constant is on the left and the y’s are on both sides. Let’s leave the constant on the left and get the variables to the right.
Try It \(\PageIndex{11}\)
Solve: \(3p−14=5p\).
Try It \(\PageIndex{12}\)
Solve: \(8m + 9 = 5m\)
Example \(\PageIndex{13}\)
Solve: \(12x = -x + 26\)
The only constant is on the right, so let the left side be the “variable” side.
Try It \(\PageIndex{14}\)
Solve: \(12j = -4j + 32\)
Try It \(\PageIndex{15}\)
Solve: \(8h = -4h + 12\)
Example \(\PageIndex{16}\): How to Solve Equations with Variables and Constants on Both Sides
Solve: \(7x + 5 = 6x + 2\)
Try It \(\PageIndex{17}\)
Solve: \(12x+8=6x+2\).
\(x=−1\)
Try It \(\PageIndex{18}\)
Solve: \(9y+4=7y+12\).
We’ll list the steps below so you can easily refer to them. But we’ll call this the ‘Beginning Strategy’ because we’ll be adding some steps later in this chapter.
BEGINNING STRATEGY FOR SOLVING EQUATIONS WITH VARIABLES AND CONSTANTS ON BOTH SIDES OF THE EQUATION.
- Choose which side will be the “variable” side—the other side will be the “constant” side.
- Collect the variable terms to the “variable” side of the equation, using the Addition or Subtraction Property of Equality.
- Collect all the constants to the other side of the equation, using the Addition or Subtraction Property of Equality.
- Make the coefficient of the variable equal 1, using the Multiplication or Division Property of Equality.
- Check the solution by substituting it into the original equation.
In Step 1, a helpful approach is to make the “variable” side the side that has the variable with the larger coefficient. This usually makes the arithmetic easier.
Example \(\PageIndex{19}\)
Solve: \(8n−4=−2n+6\).
In the first step, choose the variable side by comparing the coefficients of the variables on each side.
Try It \(\PageIndex{20}\)
Solve: \(8q - 5 = -4q + 7\)
Try It \(\PageIndex{21}\)
Solve: \(7n - 3 = n + 3\)
Example \(\PageIndex{22}\)
Solve: \(7a -3 = 13a + 7\)
Since 13>7, make the right side the “variable” side and the left side the “constant” side.
Try It \(\PageIndex{23}\)
Solve: \(2a - 2 = 6a + 18\)
Try It \(\PageIndex{24}\)
Solve: \(4k -1 = 7k + 17\)
In the last example, we could have made the left side the “variable” side, but it would have led to a negative coefficient on the variable term. (Try it!) While we could work with the negative, there is less chance of errors when working with positives. The strategy outlined above helps avoid the negatives!
To solve an equation with fractions, we just follow the steps of our strategy to get the solution!
Example \(\PageIndex{25}\)
Solve: \(\frac{4}{5}x + 6 = \frac{1}{4}x - 2\)
Since \(\frac{5}{4} > \frac{1}{4}\), make the left side the “variable” side and the right side the “constant” side.
Try It \(\PageIndex{26}\)
Solve: \(\frac{7}{8}x - 12 = -\frac{1}{8}x - 2\)
Try It \(\PageIndex{27}\)
Solve: \(\frac{7}{6}x + 11 = \frac{1}{6}y + 8\)
We will use the same strategy to find the solution for an equation with decimals.
Example \(\PageIndex{28}\)
Solve: \(7.8x+4=5.4x−8\).
Since \(7.8>5.4\), make the left side the “variable” side and the right side the “constant” side.
Try It \(\PageIndex{29}\)
Solve: \(2.8x + 12 = -1.4x - 9\)
Try It \(\PageIndex{30}\)
Solve: \(3.6y + 8 = 1.2y - 4\)
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Learn how to solve the equation 2x + 3 = 5x - 2 with the variable on both sides. We start by visualizing the equation, then isolate the variable by performing the same operations on both sides. Finally, we solve the equation to find the value of the variable. Created by Sal Khan.
You should now have the variable on one side of the equation. For example: 20 − 4 x + 4 x = 8 x + 8 + 4 x {\displaystyle 20-4x+4x=8x+8+4x} 20 = 12 x + 8 {\displaystyle 20=12x+8} 5. Move the constants to one side of the equation, if necessary. You want the variable term on one side, and the constant on the other side.
Solve an equation with variables and constants on both sides. Step 1. Choose one side to be the variable side and then the other will be the constant side. Step 2. Collect the variable terms to the variable side, using the Addition or Subtraction Property of Equality. Step 3.
Show Solution. 2 y − 7 2 y − 7 is the side containing a variable. 15 15 is the side containing only a constant. Add 7 7 to both sides. 2 y − 7 + 7 = 15 + 7 2 y − 7 + 7 = 15 + 7. Simplify. 2 y = 22 2 y = 22. The variables are now on one side and the constants on the other. Divide both sides by 2 2.
Equations with variables on both sides. Solve for f . Stuck? Review related articles/videos or use a hint. Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a nonprofit with the mission of providing a free, world-class education for anyone, anywhere.
Courses on Khan Academy are always 100% free. Start practicing—and saving your progress—now: https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:solve...
When you need to solve equations with variables on both sides of the equals sign, make sure to move all the variables to one side of the equation together. A simple saying that may help you remember this is "get all your x's to Texas." In other words, you need to move all the x terms so that they're on the same side of the equation.
Answer. Example 8.5.12: Solve: − (x + 5) = 7. Solution. Simplify each side of the equation as much as possible by distributing. The only x term is on the left side, so all variable terms are on the left side of the equation. − x − 5 = 7. Add 5 to both sides to get all constant terms on the right side of the equation.
Example 2.3.19. Solve: 8n − 4 = − 2n + 6. Solution. In the first step, choose the variable side by comparing the coefficients of the variables on each side. Since 8 > − 2, make the left side the "variable" side. We don't want variable terms on the right side—add 2n to both sides to leave only constants on the right.
Step 3: Isolate the ax term. Use the Addition Property of Equality to get the variable on one side of the equal sign and the numerical values on the other. Step 4: Isolate the variable. Use the Multiplication Property of Equality to get the variable alone on one side of the equation. Step 5: Check your solution.
Use both the Distributive Property and combining like terms to simplify and then solve algebraic equations with variables on both sides of the equation. Classifying solutions to linear equations. Some equations may have the variable on both sides of the equal sign, as in this equation: 4x−6= 2x+10 4 x − 6 = 2 x + 10.
Equations with variables on both sides involve a few more steps, so I want to make sure you understand the simpler equations first before moving forward. Let's say you have the equation below: This equation means that if you multiply a "mystery number" x by 5 and then subtract 1, the answer will be 19. To solve for x, you need to work your way ...
An equation is a mathematical statement that connects two equal expressions with an "equal to" sign. 2x + 5 = 25 - 2x, y - 6 = 10 - 3y, and 8z - 6 = 29 - z are some examples of equations having variables on both sides. The unknown values like x, y, and z found in these equations are known as variables. We can find the values of variables of an equation by rearranging the terms ...
Solution. This equation has 2a on the left and 5a on the right. Since 5 > 2, make the right side the variable side and the left side the constant side. Subtract 2a from both sides to remove the variable term from the left. 2a− 2a − 7 = 5a− 2a + 8 (8.4.30) (8.4.30) 2 a − 2 a − 7 = 5 a − 2 a + 8. Combine like terms.
Step 1: Add and subtract terms to get the variables on one side and the constants on the other. Step 2: Multiply or divide to evaluate the variable. Example: Solve 5x- 4 = 2x + 2 5 x - 4 = 2 x + 2. Step 1: Get all the variable terms to one side and the constant terms to the other side. 5x- 4 = 2x + 2 5 x - 4 = 2 x + 2.
4 years ago. because There is some way of dealing with variables on both sides called "transferring the terms", used most widely in China. ... The term actually goes to the other side of equation because on the original side the two terms cancel out or multiply to 1. So it's the same thing.
Review how to solve an equation when there is a variable on both sides of the equation. The example used in this video is 12x + 3 = 5x + 31. Visit https://...
Combine like terms on each side. Add or subtract like terms on both sides of the equation. 3. Get all variables on one side. Use addition or subtraction to move variables to one side. 4. Get all constants on the opposite side. Move constants to the opposite side using opposite operations. 5.
The golden rule for solving equations is to keep both sides of the equation balanced so that they are always equal. ... combine like terms, remove parethesis, use the order of operations. How do you solve linear equations? To solve a linear equation, get the variable on one side of the equation by using inverse operations. Show more; Why users ...
By this point in math, you would have learned how to solve equations where the variable is on both sides. Use the same techniques here to solve the inequality. -- Get the variable on only one side by using opposite operations to add/subtract one of the terms containing the variable.-- Next, If there is a constant on the same side as the ...
301 likes, 29 comments - montessoriinspiredco on February 16, 2024: "INTRO TO ALGEBRA: Hands On Equations My 8-year-old has been really eager to begin work on algebr..." Montessori Inspired | Homeschool Mom | Laura on Instagram: "INTRO TO ALGEBRA: Hands On Equations My 8-year-old has been really eager to begin work on algebra and Borenson's ...
Solve: 9x = 8x − 6 9 x = 8 x − 6. Solution. Here the variable is on both sides, but the constants only appear on the right side, so let's make the right side the "constant" side. Then the left side will be the "variable" side. We don't want any x's on the right, so subtract the 8x from both sides. Simplify.
AboutTranscript. To solve the equation (3/4)x + 2 = (3/8)x - 4, we first eliminate fractions by multiplying both sides by the least common multiple of the denominators. Then, we add or subtract terms from both sides of the equation to group the x-terms on one side and the constants on the other. Finally, we solve and check as normal.
2 likes, 0 comments - teachertwins19 on January 31, 2024: "Check out our Equations with Variables on Both Sides Weekend Escape Digital Escape Room! In each..." Teacher Twins on Instagram: "Check out our Equations with Variables on Both Sides Weekend Escape Digital Escape Room!
This topic covers: - Solving one-variable linear equations - Solving one-variable linear inequalities. If you're seeing this message, it means we're having trouble loading external resources on our website. ... Why we do the same thing to both sides: Variable on both sides (Opens a modal) Intro to equations with variables on both sides (Opens a ...