SOLUTION:Thank
In this question, we are meant to solve the system of equations using the elimination method:
15q – 4r = 62 --------- equ 1
5q + 8r = 86 ----------- equ 2
To eliminate q , we need to multiply equ 2 by 3, 15q + 24 r = 258 ------- equ 3
Then solving equ 1 and equ 3, we have:
15q – 4r = 62 ---------equ 1
15q + 24 r = 258 ------- equ 3
equ 3 - equ 2, we have:
15 q - 15 q + 24 r - ( - 4r ) = 258 - 62
24 r + 4 r = 196
28 r = 196
Dividing both sides by 28, we have:
r = 7.
Next, we put the value of r = 7, in equ 1,
15q – 4r = 62 --------- equ 1
15 q - 4 ( 7 ) = 62
15q - 28 = 62
15q = 62 + 28
15q = 90
Dividing both sides by 15, we have :
q = 6.
CONCLUSION: The value of r = 7 and q = 6.