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How do you solve $2 p + q = 1$ $2p + q = 1$ and $9 p + 3 q + 3 = 0$ $9p + 3q + 3 = 0$ using substitution?

1 Answer

Apr 8, 2016

$(p, q) = (- 2, 5)$ $(p,q)=(-2,5)$
(see below for substitution methodology)

Explanation:

Given $2 p + q = 1$ $2p+q =1$
$\to q = (1 - 2 p)$ $rarr color(red)(q)=(color(blue)(1-2p))$

Also given $9 p + 3 q + 3 = 0$ $9color(green)p+3color(brown)(q)+3=0$

Substituting $(1 - 2 p)$ $(color(blue)(1-2p))$ for $q$ $color(brown)(q)$

$XXX 9 p + 3 (1 - 2 p) + 3 = 0$ $color(white)("XXX")9color(green)(p)+3(color(blue)(1-2p))+3=0$

$XXX 3 p + 3 + 3 = 0$ $color(white)("XXX")3p+3+3=0$

$XXX p = (- 2)$ $color(white)("XXX")color(cyan)p=(color(orange)(-2))$

Substituting $(- 2)$ $(color(orange)(-2))$ for $p$ $p$ in the original equation:
$XXX 2 (- 2) + q = 1$ $color(white)("XXX")2(color(orange)(-2))+q=1$

$XXX q = 5$ $color(white)("XXX")q=5$

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