Elimination Method and Substitution method

Question

Substitution: 3x+10y=1

                       x+2y=-1

 

 

Elimination: 3/4x + 5/3y = 11

                  1/16x - 3/4y = -1

Answer 1

Substitution

 

3x + 10y = 1            eq1

x + 2y = -1              eq2

 

 

Since the coefficient of the x term in eq2 is 1, we can substitute eq2 into eq1. 

 

3(-2y - 1) + 10y = 1

 

Now we have one equation with only one variable.

 

 

-6y - 3 + 10y = 1

4y - 3 = 1

4y = 4

  y = 1

 

 

Substitute this value of y into eq1.

 

3x + 10(1) = 1

3x + 10 = 1

3x = -9

  x = -3

 

 

The solutions are

 

x = -3

y = 1

 

 

Elimination

 

(3/4)x + (5/3)y = 11           eq3

(1/16)x - (3/4)y = -1           eq4

 

 

First, we want to eliminate denominators in the equations.

 

The LCD of eq3 is 12.

 

(9x + 20y) / 12 = 132 / 12      

 

9x + 20y = 132           eq3

 

 

The LCD of eq4 is 16.

 

 

(x - 12y) / 16 = -16 / 16

 

x - 12y = -16          eq4

 

 

These are the equations we will use.

 

9x + 20y = 132               eq3

x - 12y = -16                   eq4

 

 

Multiply eq4 by 9.   Keep eq3.

 

9x + 20y = 132               eq3

9x - 108y = -144              eq4

 

 

Subtract eq4 from eq3 to eliminate the x terms.

 

128y = 276

      y = 2.16

 

 

Substitute this value of y into eq3.

 

9x + 20(2.16) = 132

9x + 43.20 = 132

9x = 88.80

  x = 9.87

 

The solutions are

 

x = 9.87

y = 2.16