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

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

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Answer 1 · 2016-08-02T20:13:10

point of intersection $\to (x, y) = (\frac{40}{41}, \frac{58}{41})$ $->(x,y)=(40/41,58/41)$

Explanation:

$Using first principles$ $color(blue)("Using first principles")$

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

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

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$Determine the value of x$ $color(blue)("Determine the value of "x)$

Consider equation (2)

Subtract $3 x$ $3x$ from both sides giving:

$5 y = 10 - 3 x$ $5y=10-3x$

Divide both sides by 5

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

$7 x - 2 y = 4 \to 7 x - 2 (\frac{10}{5} - \frac{3}{5} x) = 4$ $color(brown)(7x-2y=4)color(blue)(" "->" "7x-2(10/5-3/5x)=4$

$\Rightarrow 7 x - 4 + \frac{6}{5} x = 4$ $=>7x-4+6/5x=4$

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

Add 4 to both sides

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

$\Rightarrow x = \frac{5}{41} \times 8 = \frac{40}{41}$ $color(blue)(=>x=5/41xx8" " =" "40/41)$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$Determine the value of$ $color(blue)("Determine the value of ")$
$Using shortcuts method$ $color(brown)("Using shortcuts method")$
Substitute $x = \frac{40}{41}$ $x=40/41$ into equation(2)

$5 y + 3 x = 10 \to 5 y + 3 (\frac{40}{41}) = 10$ $color(brown)(5y+3x=10color(blue)(" "->" "5y+3(40/41)=10))$

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

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

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