Math 8

CHAPTER 2-4: Solving Equations with Variables on Both Sides of the Equation

Simplify each side of the equation first.

Then use the ADDITIVE INVERSE PROPERTY to move variables to the other side of the equation so that all variables are on the same side.

Usually, we try to move the smaller coefficient to the larger because sometimes this avoids negative coefficients,

BUT that is not always the case, and you may move to whatever side you choose!

Here’s a way to remember to use the OPPOSITE SIGN when you move terms to the other side of the equation:

OPPOSITE SIDES

use the

OPPOSITE SIGN

However if like terms are on the same side of the equation, use the sign of the coefficient that you’re given and simply combine them by using integer rules:

SAME SIDE

use the

SAME SIGN

EXAMPLE:

3y - 10 - y = -10y + 12

The 3y and -y are on the SAME SIDE, so just use the SAME SIGNS as in the equation and combine them using integer rules:

2y - 10 = -10y + 12

Now the 2y and the -10y are on OPPOSITE SIDES of the equation so use the OPPOSITE SIGN to move the -10y to the left side:

2y - 10 = -10y + 12

                                     +10   +10y

12y - 10 = 12

NOW DO THE TWO-STEP ;)

12y - 10 = 12

+ 10 +10

12y = 22

12 12

y = 11/6

FINDING THE VALUE OF AN UNKNOWN SO THAT 2 PERIMETERS OR AREAS ARE THE SAME:

This type of problem is a perfect example of using the Distributive Property with variable on both sides of an equation.

Example 4 on p. 99

You have 2 rectangles whose areas are the same. One rectangle has sides of x and 10 cm. and the other rectangle has sides of x + 3 and 6 cm.

Set up the following equation and solve:

10x = 6(x + 3)

10x = 6x + 18

4x = 18

x = 18/4 = 9/2 cm

CHECK THAT THE RECTANGLES DO HAVE THE SAME AREA IF x = 9/2 OR 4.5 cm

First rectangle: Sides are 4.5 and 10 so A = (4.5)(10) = 45 cm2

Second rectangle: Sides are 4.5 + 3 or 7.5 cm and 6 cm. so A = (7.5)(6) = 45 cm2

You have 2 rectangles whose perimeters are the same. One rectangle has sides of x and 6 cm. and the other rectangle has sides of 2x + 2 and x cm.

Set up the following equation and solve:

2x + 2(6) = 2(2x + 2) + 2x

2x + 12 = 4x + 4 + 2x

2x + 12 = 6x + 4

12 = 4x + 4

8 = 4x

x = 2

CHECK THAT THE RECTANGLES DO HAVE THE SAME PERIMETER IF x = 2 cm

First rectangle: Sides are 2 and 6 so P = 2(2) + 2(6) = 16 cm

Second rectangle: Sides are 2 and 2(2) + 2 = 6 so P = 2(2) + 2(6) = 16 cm