Lesson 1

Example 1. Use either elimination or the substitution method to solve each system of equations.

3x -2y = 7 & 2x +5y = 9 A. Using substitution method 3x -2y = 7 ( 1st eq.) 2x +5y = 9 ( 2nd eq.)

1. Choose an equation to change either in y form or x form.

For example, I choose eq. 1 & change it to x form.

3x -2y = 7 ( 1st eq.) 3x = 2y + 7 x = 3rd equation3

732

y

2. Substitute the value of x in the 2nd equation .

2( ) +5y = 9 ( 2nd eq.)

Use distributive property

Multiply both sides by 3 4y + 14 + 15y = 27

95314

34

yy

37

32

y

Multiply both sides by 3 4y + 14 + 15y = 27 by simplifying, 19y = 13 using division , y =

1913

3. Find x by substituting the value of y in the 1st equation or 3rd eq.

37)

1913(

32

x 3rd equation

By simplifying37

5726

x

Get the LCD & simplify

1953

57159

5713326

x

x

Example 1. Use either elimination or the substitution method to solve each system of equations.

3x -2y = 7 ( 1st eq.) 2x +5y = 9 ( 2nd eq.) The solution set is

1913,

1953

A. Using elimination method 3x -2y = 7 ( 1st eq.)

2x +5y = 9 ( 2nd eq.)

1. Choose a variable to eliminate. For example, I choose x. 3x -2y = 7 ( 1st eq.) 2x +5y = 9 ( 2nd eq.) To eliminate x, multiply the 1st eq. by -2 and multiply the 2nd eq. by 3. then add the 1st eq. to the 2nd eq..

-2(3x -2y = 7) ( 1st eq.) 3(2x +5y = 9) ( 2nd eq.)

-6x + 4y = -14 + 6x +15y = 27 19y = 13 y = 13 19 2. Now substitute the value of y in either equation to find the value of x.

I choose the 2nd equation. 2x +5y = 9) ( 2nd eq.)

Multiply both sides by 19 38 x + 65 = 171 38x = 171- 65 (subtract both

sides by 65)

919652

9191352

x

x

38x = 171- 65 (subtract both sides by 65)

38 x = 106 (by subtraction prop.)

X = 106 (by division prop.) 38 X = 53 (by simplifying) 19