How do you solve $3 x + 4 y = - 11$ $3x+4y=-11$ and $7 x - 5 y = 3$ $7x-5y=3$ using substitution?

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Answer 1 · 2016-07-17T03:28:18

You eliminate one variable (say $y$ $y$ ), ending up with an equation for only $x$ $x$ , leading to the solution $x = - 1, y = - 2$ $x = -1, y=-2$

You start by rewriting the first equation as

$y = - \frac{1}{4} (3 x + 11)$ $y = -1/4 (3x+11)$

Substituting this in the second equation gives

$7 x + \frac{5}{4} (3 x + 11) = 3$ $7x +5/4(3x+11) = 3$

or

$28 x + 15 x + 55 = 12$ $28x+15x+55 = 12$

Thus $43 x = - 43$ $43x = -43$ , leading to $x = - 1$ $x=-1$ ., and

$y = - \frac{1}{4} (3 \times (- 1) + 11) = - 2$ $y=-1/4 (3 xx (-1)+11) = -2$

How do you solve 3x+4y=-113x+4y=−113x+4y=-11 and 7x-5y=37x−5y=37x-5y=3 using substitution?