Solve Equations with Variables and Constants on Both Sides

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

Try it.

Solve: \(7x+8=-13.\)

Solution

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 “\(x\)”, 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.

Use the Subtraction Property of Equality.
Simplify.
Now all the variables are on the left and the constant on the right. The equation looks like those you learned to solve earlier.
Use the Division Property of Equality.
Simplify.
Check:
Let \(x=-3\).

Example

Try it.

Solve: \(8y-9=31.\)

Solution

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.

Add 9 to both sides.
Simplify.
	The variables are now on one side and the constants on the other. We continue from here as we did earlier.
Divide both sides by 8.
Simplify.
Check:
Let \(y=5\).

Solve Equations with Variables on Both Sides

What if there are variables on both sides of the equation? For equations like this, begin as we did above—choose a “variable” side and a “constant” side, and then use the subtraction and addition properties of equality to collect all variables on one side and all constants on the other side.

Example

Try it.

Solve: \(9x=8x-6.\)

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.
We succeeded in getting the variables on one side and the constants on the other, and have obtained the solution.
Check:
Let \(x=-6\).

Example

Try it.

Solve: \(5y-9=8y.\)

Solution

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.

Subtract \(5y\) from both sides.
Simplify.
We have the y’s on the right and the constants on the left. Divide both sides by 3.
Simplify.
Check:
Let \(y=-3.\)

Example

Try it.

Solve: \(12x=\text{-}x+26.\)

Solution

The only constant is on the right, so let the left side be the “variable” side.

Remove the \(-x\) from the right side by adding \(x\) to both sides.
Simplify.
All the \(x\)’s are on the left and the constants are on the right. Divide both sides by 13.
Simplify.

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.

How to Solve Equations with Variables and Constants on Both Sides

Try it.

Solve: \(7x+5=6x+2.\)

Solution

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.

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

Try it.

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.
Combine like terms.
We don’t want any constants on the left side, so add \(4\) to both sides.
Simplify.
The variable term is on the left and the constant term is on the right. To get the coefficient of \(n\) to be one, divide both sides by 10.
Simplify.
Check:
Let \(n=1\).

Example

Try it.

Solve: \(7a-3=13a+7.\)

Solution

In the first step, choose the variable side by comparing the coefficients of the variables on each side.

Since \(13>7\), make the right side the “variable” side and the left side the “constant” side.

Subtract \(7a\) from both sides to remove the variable term from the left.
Combine like terms.
Subtract \(7\) from both sides to remove the constant from the right.
Simplify.
Divide both sides by \(6\) to make \(1\) the coefficient of \(a\).
Simplify.
Check:
Let \(a=-\frac{5}{3}\).

To solve an equation with fractions, we just follow the steps of our strategy to get the solution!

Condensed — the full section is in OpenStax Elementary Algebra 2e.

Key Concepts

• Beginning Strategy for Solving an Equation with Variables and Constants on Both Sides of the Equation
  1. Choose which side will be the “variable” side—the other side will be the “constant” side.
  2. Collect the variable terms to the “variable” side of the equation, using the Addition or Subtraction Property of Equality.
  3. Collect all the constants to the other side of the equation, using the Addition or Subtraction Property of Equality.
  4. Make the coefficient of the variable equal 1, using the Multiplication or Division Property of Equality.
  5. Check the solution by substituting it into the original equation.

Solve Equations with Variables and Constants on Both Sides

Solve Equations with Constants on Both Sides

In the following exercises, solve the following equations with constants on both sides.

Try it.

\(9x-3=60\)

Try it.

\(12x-8=64\)

Solution

\(x=6\)

Try it.

\(14w+5=117\)

Try it.

\(15y+7=97\)

Solution

\(y=6\)

Try it.

\(2a+8=-28\)

Try it.

\(3m+9=-15\)

Solution

\(m=-8\)

Try it.

\(-62=8n-6\)

Try it.

\(-77=9b-5\)

Solution

\(b=-8\)

Try it.

\(35=-13y+9\)

Try it.

\(60=-21x-24\)

Solution

\(x=-4\)

Try it.

\(-12p-9=9\)

Try it.

\(-14q-2=16\)

Solution

\(q=-\frac{9}{7}\)

Solve Equations with Variables on Both Sides

In the following exercises, solve the following equations with variables on both sides.

Try it.

\(19z=18z-7\)

Try it.

\(21k=20k-11\)

Solution

\(k=-11\)

Try it.

\(9x+36=15x\)

Try it.

\(8x+27=11x\)

Solution

\(x=9\)

Try it.

\(c=-3c-20\)

Try it.

\(b=-4b-15\)

Solution

\(b=-3\)

Try it.

\(9q=44-2q\)

Try it.

\(5z=39-8z\)

Solution

\(z=3\)

Try it.

\(6y+\frac{1}{2}=5y\)

Try it.

\(4x+\frac{3}{4}=3x\)

Solution

\(x=-\frac{3}{4}\)

Try it.

\(-18a-8=-22a\)

Try it.

\(-11r-8=-7r\)

Solution

\(r=-2\)

Solve Equations with Variables and Constants on Both Sides

In the following exercises, solve the following equations with variables and constants on both sides.

Try it.

\(8x-15=7x+3\)

Try it.

\(6x-17=5x+2\)

Solution

\(x=19\)

Try it.

\(26+13d=14d+11\)

Try it.

\(21+18f=19f+14\)

Solution

\(f=7\)

Try it.

\(2p-1=4p-33\)

Try it.

\(12q-5=9q-20\)

Solution

\(q=-5\)

Try it.

\(4a+5=\text{-}a-40\)

Try it.

\(8c+7=-3c-37\)

Solution

\(c=-4\)

Try it.

\(5y-30=-5y+30\)

Try it.

\(7x-17=-8x+13\)

Solution

\(x=2\)

Try it.

\(7s+12=5+4s\)

Try it.

\(9p+14=6+4p\)

Solution

\(p=-\frac{8}{5}\)

Try it.

\(2z-6=23-z\)

Try it.

\(3y-4=12-y\)

Solution

\(y=4\)

Try it.

\(\frac{5}{3}c-3=\frac{2}{3}c-16\)

Try it.

\(\frac{7}{4}m-7=\frac{3}{4}m-13\)

Solution

\(m=-6\)

Try it.

\(8-\frac{2}{5}q=\frac{3}{5}q+6\)

Try it.

\(11-\frac{1}{5}a=\frac{4}{5}a+4\)

Solution

\(a=7\)

Try it.

\(\frac{4}{3}n+9=\frac{1}{3}n-9\)

Try it.

\(\frac{5}{4}a+15=\frac{3}{4}a-5\)

Solution

\(a=-40\)

Try it.

\(\frac{1}{4}y+7=\frac{3}{4}y-3\)

Try it.

\(\frac{3}{5}p+2=\frac{4}{5}p-1\)

Solution

\(p=15\)

Try it.

\(14n+8.25=9n+19.60\)

Try it.

\(13z+6.45=8z+23.75\)

Solution

\(z=3.46\)

Try it.

\(2.4w-100=0.8w+28\)

Try it.

\(2.7w-80=1.2w+10\)

Solution

\(w=60\)

Try it.

\(5.6r+13.1=3.5r+57.2\)

Try it.

\(6.6x-18.9=3.4x+54.7\)

Solution

\(x=23\)

Try it.

Concert tickets At a school concert the total value of tickets sold was $1506. Student tickets sold for $6 and adult tickets sold for $9. The number of adult tickets sold was 5 less than 3 times the number of student tickets. Find the number of student tickets sold, s, by solving the equation \(6s+27s-45=1506\).

Try it.

Making a fence Jovani has 150 feet of fencing to make a rectangular garden in his backyard. He wants the length to be 15 feet more than the width. Find the width, w, by solving the equation \(150=2w+30+2w\).

Solution

30 feet

Condensed — the full section is in OpenStax Elementary Algebra 2e.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

1. Simplify: \(4y-9+9.\)
   If you missed this problem, review .

   Αποκάλυψέ την.

   \(4y\)

2. Solve: \(7x+8=-13.\)

   Αποκάλυψέ την.

   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 “\(x\)”, 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.

   Use the Subtraction Property of Equality.
   Simplify.
   Now all the variables are on the left and the constant on the right. The equation looks like those you learned to solve earlier.
   Use the Division Property of Equality.
   Simplify.
   Check:
   Let \(x=-3\).

3. Solve: \(3x+4=-8.\)

   Αποκάλυψέ την.

   \(x=-4\)

4. Solve: \(5a+3=-37.\)

   Αποκάλυψέ την.

   \(a=-8\)

5. Solve: \(8y-9=31.\)

   Αποκάλυψέ την.

   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.

   Add 9 to both sides.
   Simplify.
   	The variables are now on one side and the constants on the other. We continue from here as we did earlier.
   Divide both sides by 8.
   Simplify.
   Check:
   Let \(y=5\).

6. Solve: \(5y-9=16.\)

   Αποκάλυψέ την.

   \(y=5\)

7. Solve: \(3m-8=19.\)

   Αποκάλυψέ την.

   \(m=9\)

8. Solve: \(9x=8x-6.\)

   Αποκάλυψέ την.

   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.
   We succeeded in getting the variables on one side and the constants on the other, and have obtained the solution.
   Check:
   Let \(x=-6\).

9. Solve: \(6n=5n-10.\)

   Αποκάλυψέ την.

   \(n=-10\)

10. Solve: \(-6c=-7c-1.\)

    Αποκάλυψέ την.

    \(c=-1\)

11. Solve: \(5y-9=8y.\)

    Αποκάλυψέ την.

    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.

    Subtract \(5y\) from both sides.
    Simplify.
    We have the y’s on the right and the constants on the left. Divide both sides by 3.
    Simplify.
    Check:
    Let \(y=-3.\)

12. Solve: \(3p-14=5p.\)

    Αποκάλυψέ την.

    \(p=-7\)

13. Solve: \(8m+9=5m.\)

    Αποκάλυψέ την.

    \(m=-3\)

14. Solve: \(12x=\text{-}x+26.\)

    Αποκάλυψέ την.

    The only constant is on the right, so let the left side be the “variable” side.

    Remove the \(-x\) from the right side by adding \(x\) to both sides.
    Simplify.
    All the \(x\)’s are on the left and the constants are on the right. Divide both sides by 13.
    Simplify.

15. Solve: \(12j=-4j+32.\)

    Αποκάλυψέ την.

    \(j=2\)

16. Solve: \(8h=-4h+12.\)

    Αποκάλυψέ την.

    \(h=1\)

17. Solve: \(7x+5=6x+2.\)

18. Solve: \(12x+8=6x+2.\)

    Αποκάλυψέ την.

    \(x=-1\)

19. Solve: \(9y+4=7y+12.\)

    Αποκάλυψέ την.

    \(y=4\)

20. Solve: \(8n-4=-2n+6.\)

    Αποκάλυψέ την.

    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.
    Combine like terms.
    We don’t want any constants on the left side, so add \(4\) to both sides.
    Simplify.
    The variable term is on the left and the constant term is on the right. To get the coefficient of \(n\) to be one, divide both sides by 10.
    Simplify.
    Check:
    Let \(n=1\).

21. Solve: \(8q-5=-4q+7.\)

    Αποκάλυψέ την.

    \(q=1\)

22. Solve: \(7n-3=n+3.\)

    Αποκάλυψέ την.

    \(n=1\)

23. Solve: \(7a-3=13a+7.\)

    Αποκάλυψέ την.

    In the first step, choose the variable side by comparing the coefficients of the variables on each side.

    Since \(13>7\), make the right side the “variable” side and the left side the “constant” side.

    Subtract \(7a\) from both sides to remove the variable term from the left.
    Combine like terms.
    Subtract \(7\) from both sides to remove the constant from the right.
    Simplify.
    Divide both sides by \(6\) to make \(1\) the coefficient of \(a\).
    Simplify.
    Check:
    Let \(a=-\frac{5}{3}\).

24. Solve: \(2a-2=6a+18.\)

    Αποκάλυψέ την.

    \(a=-5\)

25. Solve: \(4k-1=7k+17.\)

    Αποκάλυψέ την.

    \(k=-6\)

26. Solve: \(\frac{5}{4}x+6=\frac{1}{4}x-2.\)

    Αποκάλυψέ την.

    Since \(\frac{5}{4}>\frac{1}{4}\), make the left side the “variable” side and the right side the “constant” side.

    Subtract \(\frac{1}{4}x\) from both sides.
    Combine like terms.
    Subtract \(6\) from both sides.
    Simplify.
    \(\begin{array}{llllll}\text{Check:} & & & \frac{5}{4}x+6 & = & \frac{1}{4}x-2 \\ \text{Let}\ x=-8. & & & \frac{5}{4}(-8)+6 & \overset{?}{=} & \frac{1}{4}(-8)-2 \\ & & & -10+6 & \overset{?}{=} & -2-2 \\ & & & -4 & = & -4✓\end{array}\)

27. Solve: \(\frac{7}{8}x-12=-\frac{1}{8}x-2.\)

    Αποκάλυψέ την.

    \(x=10\)

28. Solve: \(\frac{7}{6}y+11=\frac{1}{6}y+8.\)

    Αποκάλυψέ την.

    \(y=-3\)

29. Solve: \(7.8x+4=5.4x-8.\)

    Αποκάλυψέ την.

    Since \(7.8>5.4\), make the left side the “variable” side and the right side the “constant” side.

    Subtract \(5.4x\) from both sides.
    Combine like terms.
    Subtract \(4\) from both sides.
    Simplify.
    Use the Division Propery of Equality.
    Simplify.
    Check:
    Let \(x=-5\).

30. Solve: \(2.8x+12=-1.4x-9.\)

    Αποκάλυψέ την.

    \(x=-5\)

31. Solve: \(3.6y+8=1.2y-4.\)

    Αποκάλυψέ την.

    \(y=-5\)

32. \(12x-8=64\)

    Αποκάλυψέ την.

    \(x=6\)

33. \(14w+5=117\)

34. \(15y+7=97\)

    Αποκάλυψέ την.

    \(y=6\)

35. \(2a+8=-28\)

36. \(3m+9=-15\)

    Αποκάλυψέ την.

    \(m=-8\)

37. \(-62=8n-6\)

38. \(-77=9b-5\)

    Αποκάλυψέ την.

    \(b=-8\)

39. \(35=-13y+9\)

40. \(60=-21x-24\)

    Αποκάλυψέ την.

    \(x=-4\)

Symbols used here

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\neq

not equal
The two sides are different.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

i

imaginary unit
i² = −1.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

How to: Solve Equations with Variables and Constants on Both Sides

1. Solve an equation with constants on both sides
2. Solve an equation with variables on both sides
3. Solve an equation with variables and constants on both sides
4. Choose which side will be the “variable” side—the other side will be the “constant” side.
5. Collect the variable terms to the “variable” side of the equation, using the Addition or Subtraction Property of Equality.
6. Collect all the constants to the other side of the equation, using the Addition or Subtraction Property of Equality.
7. Make the coefficient of the variable equal 1, using the Multiplication or Division Property of Equality.
8. Check the solution by substituting it into the original equation.

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

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Parts of this page are adapted from OpenStax Elementary Algebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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