Solve the following pairs of linear equations by substitution and elimination methods:
2x -5y+4,
3x-2y+16=0
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Answers
Answer:
x = -72/11 and y = -20/11
Step-by-step explanation:
2x - 5y = -4 ----------------(1)
3x - 2y = -16 --------------(2)
by substitution method
from eq 1
x = (-4 + 5y)/2
put x in eq. 2
3x - 2y = -16
3{(-4+5y)/2} - 2y = -16
(-12 + 15y)/2 -2y = -16
(-12 + 15y - 4y)/2 = -16
-12 + 15y -4y = -16 × 2
-12 + 11y = -32
11y = -32 + 12
11y = -20
y = -20/11
put y in eq 2
3x - 2y = -16
3x - 2(-20/11) = -16
3x + 40/11 = -16
3x = -16 - 40/11
3x = (-176 - 40)/11
3x = -216/11
x = -216/(3 × 11)
x = -72/11
by elimination method
2x - 5y = -4 ---------(1)
3x - 2y = -16 ----------(2)
multiply by 3 in eq 1 and 2 in eq 2
3(2x - 5y = -4)
6x - 15y = -12 -----------(3)
2(3x - 2y = -16)
6x - 4y = -32 ------------(4)
on subtract eq. 3 and 4
6x - 4y = -32
6x - 15y = -12
- + +
0 + 11y = -20
y = -20/11
put y in eq 1
2x - 5y = -4
2x - 5(-20/11) = -4
2x + 100/11 = -4
2x = -4 - 100/11
2x = (-44 - 100)/11
2x = -144/11
x = -144/(11 × 2)
x = -72/11