# How do you solve 3x+y=2 and 2x+3y=13 using substitution?

Mar 4, 2016

$\text{Point of intersection "(x,y)" "->" } \left(- 1 , 5\right)$

#### Explanation:

Given:
$3 x + y = 2$............................(1)
$2 x + 3 y = 13$.......................(2)
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Consider equation (1)

Subtract $3 x$ form both sides

$3 x - 3 x + y = 2 - 3 x$

$0 + y = - 3 x + 2 \text{ } \ldots \ldots \ldots \ldots \ldots \ldots \left({1}_{a}\right)$

Using equation ${1}_{a}$ substitute for $y$ in equation (2)

$2 x + 3 \left(- 3 x + 2\right) = 13. \ldots \ldots \ldots \ldots \ldots \left({2}_{a}\right)$

$2 x - 9 x + 6 = 13$

$- 7 x + 6 = 13$

Subtract 6 from both sides

$- 7 x + 6 - 6 = 13 - 6$

$- 7 x + 0 = 7$

Divide bot sides by -7

$\frac{- 7}{- 7} \times x = \frac{+ 7}{- 7}$

$\left(+ 1\right) \times x = - 1$

$\textcolor{red}{x = - 1}$
'~~~~~~~~~~~~~~~~~~~~~~~~~
Substitute $x = - 1$ into equation (1)

$3 x + y = 2 \text{ "->" } 3 \left(- 1\right) + y = 2$

$- 3 + y = 2$

$- 3 + 3 + y = 2 + 3$
$\textcolor{red}{y = 5}$
write equation (1) as: $y = - 3 x + 2$
write equation (2) as: $y = - \frac{2}{3} x + \frac{13}{3}$