How do you solve $3 x + 2 y = 7$ $3x + 2y = 7$ and $x - y + 3 = 0$ $x - y + 3 = 0$ using substitution?

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Answer 1 · 2017-03-31T09:55:32

$y = 3 \frac{1}{5}, x = \frac{1}{5}$ $y=3 1/5,x=1/5$

$:.3x+2y=7$ ---------- (1)

$:.x-y+3=0$

$:.x-y=-3$ ---------(2)

----(2) $xx 3$ :

$:.3x-3y=-9$ --------(3)

----- $(1)-(3)$ :

$:.3x+2y=7$
$:.3x-3y=-9$
$:. 5y=16$

$:.y=16/5$

$:.y=3 1/5$

substitute $y=3 1/5$ in $(2)$

$:.x-(3 1/5)=-3$

$:.x=-3+3 1/5$

$:.x=1/5$

substitute $y=16/5,x=1/5$ in $(1)$

$:.3(1/5)+2(16/5)=7$

$:.3/5+32/5=7$

$:.35/5=7$

$:.7=7$

How do you solve 3x + 2y = 73x+2y=73x + 2y = 7 and x - y + 3 = 0 x−y+3=0x - y + 3 = 0 using substitution?