How do you solve 7x - 2y = 4 and 5y + 3x = 10?

Aug 2, 2016

point of intersection$\to \left(x , y\right) = \left(\frac{40}{41} , \frac{58}{41}\right)$

Explanation:

$\textcolor{b l u e}{\text{Using first principles}}$

There are various ways of solving this question type. What follows is just one of them.

Given:
$7 x - 2 y = 4$ ...............................Equation(1)
$5 y + 3 x = 10$ ............................Equation(2)

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$\textcolor{b l u e}{\text{Determine the value of } x}$

Consider equation (2)

Subtract $3 x$ from both sides giving:

$5 y = 10 - 3 x$

Divide both sides by 5

$y = \frac{10}{5} - \frac{3}{5} x$ ..............................Equation(3)
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Using equation(3) substitute for y in equation (1)

color(brown)(7x-2y=4)color(blue)(" "->" "7x-2(10/5-3/5x)=4

$\implies 7 x - 4 + \frac{6}{5} x = 4$

$\frac{41}{5} x - 4 = 4$

Add 4 to both sides

$\frac{41}{5} x = 8$
Multiply both sides by $\frac{5}{41}$

$\textcolor{b l u e}{\implies x = \frac{5}{41} \times 8 \text{ " =" } \frac{40}{41}}$

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$\textcolor{b l u e}{\text{Determine the value of }}$
$\textcolor{b r o w n}{\text{Using shortcuts method}}$
Substitute $x = \frac{40}{41}$ into equation(2)

$\textcolor{b r o w n}{5 y + 3 x = 10 \textcolor{b l u e}{\text{ "->" } 5 y + 3 \left(\frac{40}{41}\right) = 10}}$

$y = \frac{58}{41}$ 