# Question #80df0

May 23, 2017

$\left(1 , 3\right)$

#### Explanation:

$\textcolor{red}{- 2 x} + 5 y = 13 \to \left(1\right)$

$\textcolor{red}{16 x} + 3 y = 25 \to \left(2\right)$

$\textcolor{b l u e}{\text{solving by elimination}}$ means attempting to eliminate one of the variables x or y from an equation leaving the equation with only one variable which we can solve.

$\text{note that multiplying " color(red)(-2x)" in (1) by 8 gives}$
$\textcolor{red}{- 16 x} \text{ and adding it to " color(red)(16x)" in (2) will }$
$\textcolor{b l u e}{\text{eliminate"" the x-term}}$

$\text{multiply ALL terms in " (1)" by 8}$

$\Rightarrow \textcolor{red}{- 16 x} + 40 y = 104 \to \left(3\right)$

$\text{adding equations " (2)" and " (3)" term by term gives}$

$\left(16 x - 16 x\right) + \left(3 y + 40 y\right) = \left(25 + 104\right)$

$\Rightarrow 43 y = 129 \leftarrow \textcolor{red}{\text{ variable x is eliminated}}$

$\text{divide both sides by 43}$

$\frac{\cancel{43} y}{\cancel{43}} = \frac{129}{43}$

$\Rightarrow y = 3$

$\text{substitute this value in either " (1)" or } \left(2\right)$

$16 x + 9 = 25 \leftarrow \text{ substituting in } \left(2\right)$

$\text{subtract 9 from both sides}$

$16 x \cancel{+ 9} \cancel{- 9} = 25 - 9$

$\Rightarrow 16 x = 16 \Rightarrow x = 1$

$\left(1 , 3\right) \text{ is the point of intersection of the 2 linear equations}$ graph{(y-2/5x-13/5)(y+16/3x-25/3)=0 [-10, 10, -5, 5]}