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How do you solve the following system: $4 x + 3 y = - 1, x - y = 3$ $4x+3y= -1 , x-y=3$ ?

1 Answer

Apr 3, 2018

$x = \frac{8}{7} or 1 \frac{1}{7} xxxx y = - \frac{13}{7} or - 1 \frac{6}{7}$ $x=8/7" or " 1 1/7color(white)("xxxx")y=-13/7" or " -1 6/7$

Explanation:

Given
[1] $XXX 4 x + 3 y = - 1$ $color(white)("XXX")4x+3y=-1$
[2] $XXX x - y = 3$ $color(white)("XXX")x-y=3$

Re-arranging [2]
[3] $XXX x = y + 3$ $color(white)("XXX")x=y+3$

Substituting $(y + 3)$ $(y+3)$ for $x$ $x$ in [1]
[4] $XXX 4 (y + 3) + 3 y = - 1$ $color(white)("XXX")4(y+3)+3y=-1$

Expanding and simplifying the lefts side of [4]
[5] $XXX 7 y + 12 = - 1$ $color(white)("XXX")7y+12=-1$

Subtracting $12$ $12$ from both sides of [5]
[6] $XXX 7 y = - 13$ $color(white)("XXX")7y=-13$

Dividing both sides of [6] by 7
[8] $XXX y = - \frac{13}{7}$ $color(white)("XXX")y=-13/7$

Substituting $(- \frac{13}{7})$ $(-13/7)$ for $y$ $y$ in [3]
[9] $XXX x = - \frac{13}{7} + 3 = \frac{- 13 + 21}{7} = \frac{8}{7}$ $color(white)("XXX")x=-13/7+3=(-13+21)/7=8/7$

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