Question

solve the following pairs of linear equation using method of elimination by substitution
11.
3x + 2y = 11
2x – 3y + 10 = 0
12.
2x – 3y + 6 = 0
2x + 3y - 18 = 0​

Answer 1

Answer:

11. x =1, y=4

12. x =3, y = 2.

Step-by-step explanation:

3x+2y=11

2x+3y+10=0

3x=11-2y (making x the subject)

x=11-2y/3

2(11-2y/3) - 3y + 10=0 (substituting x in the second equation to find y)

$(22 - 4y \div 3) - (3y \times 3 \div 1 \times 3) + (10 \times 3 \div 1 \times 3) = 0$

$(22 - 4y - 9y + 30) \div 3 = 0$

$(22 + 30 - 4y - 9y) \div 3 = 0$

$(52 - 13y) = 0$

52= 13y

52/13= y

4=y

So I got the value of y, now I am going to use either of the equations to substitute the value to find x.

3x+2(4)= 11 ( I substituted the value of y here)

3x+ 8=11

3x= 11-8

3x=3

x= 3/3

×=1

12.

$2x - 3y + 6 = 0$

$2x + 3y - 18 = 0$

The y here can be subtracted since they have the same coefficient with addition and subtraction in front of them.

2x+2x= 4x

6-18= -12

4x-12= 0

4x= 12

x= 12/4

x=3

The value of x now replaces in either of the equations.

I choose the first equation.

2(3) - 3y + 6 = 0

6 + 6 - 6y=0

12 - 6y = 0

-6y = -12

y= -12 ÷ -6

the substactions cut off each other.

so,

y= 2