How do you solve the following system?: $y = \frac{3}{4} x + 5, x + 2 y = 3$ $y = 3/4x + 5 , x + 2y = 3$

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Answer 1 · 2016-02-27T08:58:05

$x = - \frac{14}{5}$ $x=-14/5$ and $y = \frac{29}{10}$ $y=29/10$

From the given equations
$y = \frac{3 x}{4} + 5$ $y=(3x)/4+5$ and $x + 2 y = 3$ $x+2y=3$

Using the second equation $x + 2 y = 3$ $x+2y=3$
Let $x = 3 - 2 y$ $x=3-2y$ then substitute in the first equation

first equation becomes

$y = \frac{3 (3 - 2 y)}{4} + 5$ $y=(3(3-2y))/4+5$

multiply both sides by of the equation by 4 now to get rid of any fraction

$4 y = 4 \cdot \frac{3 (3 - 2 y)}{4} + 4 \cdot 5$ $4y=4*(3(3-2y))/4+4*5$

$4y=cancel4*(3(3-2y))/cancel4+4*5$

$4y=9-6y+20$

$4y+6y=9+20$

$10y=29$
divide both sides by 10

$(10y)/10=29/10$

$(cancel10y)/cancel10=29/10$

$y=29/10$

Solve now for the value of x using the second equation

$x=3-2y$ and $y=29/10$

$x=3-2*(29/10)$
$x=3-29/5$
$x=(15-29)/5$

$x=-14/5$

God bless...I hope the explanation is useful..

How do you solve the following system?: y = 3/4x + 5 , x + 2y = 3 y=34x+5,x+2y=3 y = 3/4x + 5 , x + 2y = 3