How do you solve 3/4x + 1/3y = 1 and x - y = 10?

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Answer 1 · 2018-07-14T11:52:50

$x = 10 + y$ $x=10+y$ substitute this into the first equation

$\frac{3}{4} (10 + y) + \frac{1}{3} y = 1$ $3/4(10+y)+1/3y=1$

$\frac{30}{4} + \frac{3}{4} y + \frac{1}{3} y = 1$ $30/4+3/4y+1/3y=1$

Multiply by 12 to remove the fractions

$90 + 9 y + 4 y = 12$ $90+9y+4y=12$

$90 + 13 y = 12$ $90+13y=12$

$13 y = - 78$ $13y=-78$

$y = - 6$ $y=-6$

Put #y=-6 into equation 2

$x - - 6 = 10$ $x- -6=10$

$x = 4$ $x=4$

Answer 2 · 2018-07-14T14:29:47

$x = 4, y = - 6$ $x=4,y=-6$

$:.3/4x+1/3y=1---(1)$

$:.x-y=10----(2)$

$:.(2) xx 3/4$

$:.3/4x-3/4y=15/2----(3)$

$:.(1)-(3)$

$:.13/12y=-13/2$

$:.y=-13/2-:13/12$

$:.y=-cancel 13^1/cancel2^1 xx cancel12^6/cancel 13^1$

$:.y=-6$

$"substitute"$ y=-6 $" in" (2)$

$:.x-(-6)=10$

$:.x+6=10$

$:.x=10-6$

$:.x=4$

~~~~~~~~

check:-

$"substitute"$ y=-6 $and$ x=3 $"in"(1)$

$:.3/4(4)+1/3(-6)=1$

$:.3-2=1$

$:.1=1$