# SAT MATH - SOLVING EQUATIONS WITH VARIABLES ON BOTH SIDES

Some equations have variables on both sides. To solve such equations, first use the Addition or Subtraction  Property of Equality to write an equivalent equation that has all of the variables on one side. Then use the  Multiplication or Division Property of Equality to simplify the equation if necessary. When solving equations that   contain grouping symbols, use the Distributive Property to remove the grouping symbols.

Example 1 :

Solve for k :

5 - 3(k + 1) = -k

Solution :

5 - 3(k + 1) = -k

5 - 3k - 3 = -k

-3k + 2 = -k

-2k + 2 = 0

Subtract 2 from both sides.

-2k = -2

Divide both sides by -2.

k = 1

Example 2 :

Solve for x :

Solution :

7x - 24 = 18 + x

6x - 24 = 18

6x = 42

x = 7

Example 3 :

7n + 3 = 2n - 12

Given the equation above, find the value of (-n + 3).

Solution:

7n + 3 = 2n - 12

5n + 3 = -12

5n = -15

n = -5

Multiply both sides by -1.

-n = 5

-n + 3 = 8

Example 4 :

7(h - 5) - 3h = (³⁄₂)h

Given the equation above, find the value of ()h.

Solution:

7(h - 5) - 3h = (³⁄₂)h

Multiply both sides by 2.

2[7(h - 5) - 3h] = 2(³⁄₂)h

14(h - 5) - 6h = 3h

14h - 70 - 6h = 3h

8h - 70 = 3h

5h - 70 = 0

5h = 70

h = 14

Multiply both sides by .

()h = ⅐ ⋅ 14

()h = 2

Example 5 :

Solution:

Substitute s = 2.

Example 6 :

⁽⁹ ⁻ ²ᵏ⁾⁄₃ = k - 2

Given the equation above, find the value of k.

Solution :

⁽⁹ ⁻ ²ᵏ⁾⁄₃ = k - 2

Multiply both sides by 3.

9 - 2k = 3(k - 2)

9 - 2k = 3k - 6

Subtract 3k from both sides.

9 - 5k = -6

Subtract 9 from both sides.

-5k = -15

Divide both sides by -5.

k = 3

Example 7 :

Four times the sum of three and a number equals nine less than the number.

(a) Write an equation for the problem.

(b) Then solve the equation.

Solution :

Part (a) :

Let x be the number.

Given : Four times the sum of three and a number equals nine less than the number.

4(x + 3) = x - 9

Part (b) :

Solve the equation in part (a).

4(x + 3) = x - 9

4x + 12 = x - 9

Subtract x from both sides.

3x + 12 = -9

Subtract 12 from both sides.

3x = -21

Divide both sides by 3.

x = -7

Example 8 :

A shirt costs \$22 more than one half the cost of a pair of pants. If the cost of a shirt is \$48, find the cost of a pair of pants.

Solution :

Let s be the cost of a shirt and p be the cost of a pair of pants.

Given :  The shirt costs \$22 more than one half the cost of a pair of pants, you can set up the following equation.

Given : The cost of a shirt is \$48.

Substitute s = 48.

Therefore, the cost of a pair of pants is \$52.

Example 9 :

Twice a number n, increased by 11 is the same as six times the number decreased by 9. What is the value of n?

Solution :

From the given information,

2n + 11 = 6n - 9

Subtract 6n from from both sides.

-4n + 11 = -9

Subtract 11 from both sides.

-4n = -20

Divide both sides by -4.

n = 5

Example 10 :

One half of a number increased by 3 is five less than two thirds of the number. Find the number.

Solution :

Let x be the number.

From the given information,

Example 11 :

Four times the greatest of three consecutive odd integers exceeds three times the least by 31. What is the greatest of the three consecutive odd integers?

Solution :

Let x be odd integer. Then the three consecutive odd integers are

x, (x + 2), (x + 4)

Given : Four times the greatest of three consecutive odd integers exceeds three times the least by 31.

So, 4(x + 6) eceeds the 3x by 31.

For example, if 3x = 100, 4(x + 4) = 131.

﻿131 = 100 + 31

Then, we have

4(x + 4) = 3x + 31

4x + 16 = 3x + 31

x + 16 = 31

x = 15

The greatest of the three consecutive odd integers :

= x + 4

= 15 + 4

= 19

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