Solve Equations by Collecting Variables on Either Side | Grade 8 Math | LevelKite

You already know how to find the number $x$ that makes $2x + 2 = 10$ true. Here are the words for that work. A variable is a letter, such as $x$ , that stands for a number. An expression, such as $2x + 2$ , combines numbers, variables and operations. An equation is a statement that two expressions are equal, and a solution is a value of the variable that makes the equation true.

In $2x + 2 = 10$ , the variable is on one side only. In this lesson, it appears on both sides, as in $5x + 2 = 3x + 10$ .

Both sides use the same unknown number $x$ . Here $x$ may be any real number (any number on the number line) unless a problem says otherwise. The equation asks one question: for which values of $x$ do the left side and the right side have the same value? The table below tests a few values.

Testing values of x in 5x + 2 = 3x + 10

Value of x	Left side: 5x + 2	Right side: 3x + 10	Are the sides equal?
$2$	$5(2) + 2 = 12$	$3(2) + 10 = 16$	No
$3$	$5(3) + 2 = 17$	$3(3) + 10 = 19$	No
$4$	$5(4) + 2 = 22$	$3(4) + 10 = 22$	Yes
$5$	$5(5) + 2 = 27$	$3(5) + 10 = 25$	No

The table shows that $x = 4$ is a solution: both sides equal $22$ . But testing values is slow, and a table cannot tell you whether an untested number, such as a fraction, also works. You need a method that settles the question.

Here is the idea behind the method. For any value of $x$ , $3x$ is just a number. So when you subtract $3x$ from both sides of $5x + 2 = 3x + 10$ , you take the same number away from each side, and the sides stay equal whenever they were equal before. You can undo the step by adding $3x$ back to both sides, so you gain no solution and lose none.

Since $5x - 3x = 2x$ and $3x - 3x = 0$ , the new equation is $2x + 2 = 10$ . It is true for exactly the same values of $x$ as the original equation. Two equations with exactly the same solutions are equivalent equations. Better still, the variable is now on one side only: this is a multi-step equation of the kind you already know how to solve.

Key concept

Collect the variable on one side; either side works

To solve an equation with the variable on both sides, add or subtract a variable term on both sides so that the variable is on one side only. Then solve the multi-step equation that remains, and check your answer in the original equation.

You may collect the variable on the left or on the right. Each step adds or subtracts the same term on both sides, or divides both sides by the same number that is not zero. Steps like these always give an equivalent equation, so both routes must end with the same solutions as the original equation.

Worked example

Collect the variable on the left

Solve $5x + 2 = 3x + 10$ by collecting the variable terms on the left side.

Subtract 3x from both sides to clear the variable term from the right side. This is the same operation on both sides, so the new equation is equivalent to the original. Then combine like terms on each side: 5x - 3x is 2x, and 3x - 3x is 0.
$5x + 2 - 3x = 3x + 10 - 3x \quad\text{so}\quad 2x + 2 = 10$
The variable is now on the left side only. Subtract 2 from both sides so that the variable term stands alone. Subtracting the same number from both sides keeps the equation equivalent.
$2x + 2 - 2 = 10 - 2 \quad\text{so}\quad 2x = 8$
Divide both sides by 2, the coefficient of x. Dividing both sides by the same number keeps the equation equivalent because 2 is not zero.
$\frac{2x}{2} = \frac{8}{2} \quad\text{so}\quad x = 4$
Check the value in the original equation, because a check there also catches a slip in any later line. Replace x with 4 on each side and compare the results.
$5(4) + 2 = 20 + 2 = 22 \quad\text{and}\quad 3(4) + 10 = 12 + 10 = 22$

Answer

The solution is $x = 4$ . Both sides of the original equation equal $22$ when $x = 4$ , and because every step gave an equivalent equation, no other value of $x$ works.

Worked example

The same equation, collected on the right

Solve $5x + 2 = 3x + 10$ again. This time, collect the variable terms on the right side.

Subtract 5x from both sides to clear the variable term from the left side. It is the same operation on both sides, so the new equation is equivalent to the original. Combine like terms: 5x - 5x is 0 on the left, and 3x - 5x is -2x on the right.
$5x + 2 - 5x = 3x + 10 - 5x \quad\text{so}\quad 2 = -2x + 10$
Subtract 10 from both sides, again the same operation on each side, so that the variable term stands alone on the right.
$2 - 10 = -2x + 10 - 10 \quad\text{so}\quad -8 = -2x$
Divide both sides by -2, the coefficient of x. This keeps the equation equivalent because -2 is not zero. A negative number divided by a negative number is positive, so -8 divided by -2 is 4.
$\frac{-8}{-2} = \frac{-2x}{-2} \quad\text{so}\quad 4 = x$
The statement 4 = x says the same thing as x = 4. Check the value in the original equation: each side must come out the same.
$5(4) + 2 = 22 \quad\text{and}\quad 3(4) + 10 = 22$

Answer

The solution is $x = 4$ again. Collecting the variable terms on the right gives the same solution as collecting them on the left.

Both routes ended at $x = 4$ , and that is no accident.

Every step in both examples was one of these: adding or subtracting the same term on both sides, or dividing both sides by the same number that is not zero. You can undo each such step, so each one gives an equivalent equation: one with exactly the same solutions as the equation before it. Combining like terms only rewrites one side as an equivalent expression, so it does not change the solutions either.

Both routes start from the same equation, $5x + 2 = 3x + 10$ . So the last line of each route, $x = 4$ or $4 = x$ , must have exactly the same solutions as that starting equation. The two routes cannot disagree.

The routes even meet along the way. Multiply both sides of $2x = 8$ by $-1$ , which is not zero, and you get $-2x = -8$ : the second route's equation with its sides swapped.

The next table tests every equation from both routes. Each one is true when $x = 4$ and false when $x = 3$ , a value that is not a solution.

Every equation from both routes, tested at x = 4 and at x = 3

Equation	Where it comes from	At x = 4	At x = 3
$5x + 2 = 3x + 10$	The original equation	$22 = 22$ , true	$17 = 19$ , false
$2x + 2 = 10$	Collecting on the left	$10 = 10$ , true	$8 = 10$ , false
$2x = 8$	Collecting on the left	$8 = 8$ , true	$6 = 8$ , false
$2 = -2x + 10$	Collecting on the right	$2 = 2$ , true	$2 = 4$ , false
$-8 = -2x$	Collecting on the right	$-8 = -8$ , true	$-8 = -6$ , false
$x = 4$	The last line of both routes	$4 = 4$ , true	$3 = 4$ , false

A common mistake: adding across the equals sign. In $5x + 2 = 3x + 10$ , you might gather the variable terms by adding them, $5x + 3x = 8x$ , and write $8x + 2 = 10$ . That gives $8x = 8$ , so $x = 1$ . Check in the original equation: the left side is $5(1) + 2 = 7$ , but the right side is $3(1) + 10 = 13$ . The sides are not equal, so $1$ is not a solution.

Why it fails: this step takes $3x$ away from the right side but adds $3x$ to the left side. Those are different operations on the two sides, so the new equation is not equivalent to the original. You can combine like terms only when they are on the same side. To clear $3x$ from the right, subtract $3x$ from both sides. The left side becomes $5x - 3x = 2x$ , not $8x$ .

Worked example

Choose the easier side

Solve $3 - 2x = 4x + 18$ .

Either side works, so pick the easier one. The left side takes away 2x, so add 2x to both sides to clear it. This is the same operation on both sides, so the new equation is equivalent to the original. The coefficient that remains, 4 + 2 = 6, is positive.
$3 - 2x + 2x = 4x + 18 + 2x \quad\text{so}\quad 3 = 6x + 18$
Subtract 18 from both sides, the same operation on each side, so that the variable term stands alone.
$3 - 18 = 6x + 18 - 18 \quad\text{so}\quad -15 = 6x$
Divide both sides by 6, which is not zero. Simplify the fraction by dividing 15 and 6 by 3. The solution is not a whole number, so keep it as an exact fraction.
$\frac{-15}{6} = \frac{6x}{6} \quad\text{so}\quad x = -\frac{15}{6} = -\frac{5}{2}$
Check the value in the original equation by replacing x with -5/2 on each side. On the left, 2 times -5/2 is -5, and subtracting -5 is the same as adding 5.
$3 - 2\left(-\tfrac{5}{2}\right) = 3 - (-5) = 3 + 5 = 8 \quad\text{and}\quad 4\left(-\tfrac{5}{2}\right) + 18 = -10 + 18 = 8$
Confirm that the other side gives the same solution. Subtract 4x from both sides, subtract 3 from both sides, then divide both sides by -6, which is not zero. You track more negative signs, but you reach the same result.
$3 - 6x = 18 \quad\text{so}\quad -6x = 15 \quad\text{so}\quad x = \frac{15}{-6} = -\frac{5}{2}$

Answer

The solution is $x = -\tfrac{5}{2}$ , which is exactly $-2.5$ . Both sides of the original equation equal $8$ for this value, and collecting on either side leads to it.

Takeaway. When the variable appears on both sides, add or subtract a variable term on both sides so that the variable is on one side only. Then solve the multi-step equation that remains, and check your answer in the original equation.

You can collect on the left or on the right. Every step adds or subtracts the same term on both sides, or divides both sides by the same number that is not zero, so every line has exactly the same solutions as the equation you started with. Pick the side that keeps the arithmetic simple.

Next, you will solve equations with variables on both sides that have parentheses, using the distributive property to remove them. You will also meet fraction and decimal coefficients, and you will find and correct errors in solutions.