How do you use substitution to solve b+g=112 and b=2g+16?
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Answered by
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Solution:
b + g = 112
b = 112 - g -------- (a)
And,
- b = 2g + 16 ------- (b)
Subtracting equation (b) from (a), we get,
- b - b = 112 - g - (2g + 16)
- 0 = 112 - g - 2g - 16
- 0 = 96 - 3g
- 3g = 96
- g = 96/3
- g = 32
Putting the value of g in equation (a), we get,
- b = 112 - g
- b = 112 - 32
- b = 80
Hence, the value of g and b is 32 and 80 respectively.
Answered by
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b+g=112 -> eq. 1
b=2g+16
b-2g=16 -> eq. 2
on multiplying eq. 1 by 2.
2 (b+g)=2×112
2b+2g=224 -> eq. 3
on adding eq. 2 and eq.3
2b+2g=224
+b-2g=16
2g cancel.
3b=240
b=240/3
b=80
on substituting b=80 in eq. 1
b+g = 112
80+g=112
g=112-80
g=32
answer: b=80
g=32
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