How do you solve the following system?: $4 x - 3 y = 15, - 6 x + 3 y = 17$ $4x-3y=15 , -6x+3y=17$

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Answer 1 · 2016-01-30T07:45:38

$(x, y) = (- 16, - \frac{79}{3})$ $(x,y)=(-16,-79/3)$

Solve by elimination

$4 x - 3 y = 15$ $4x-3y=15$

$- 6 x + 3 y = 17$ $-6x+3y=17$

If we loo carefully we can see that we can eliminate $3 y$ $3y$ in the second equation by $- 3 y$ $-3y$ in the first equation.

Add the equations

$\to (4 x - 3 y = 15) + (- 6 x + 3 y = 17)$ $rarr(4x-3y=15)+(-6x+3y=17)$

$\to - 2 x = 32$ $rarr-2x=32$

$\to x = \frac{32}{- 2} = - 16$ $rarrx=32/-2=-16$

So,substitute the value of $x$ $x$ to the first equation:

$\to 4 (- 16) - 3 y = 15$ $rarr4(-16)-3y=15$

$\to - 64 - 3 y = 15$ $rarr-64-3y=15$

$\to - 3 y = 15 + 64$ $rarr-3y=15+64$

$\to - 3 y = 79$ $rarr-3y=79$

$\to y = - \frac{79}{3}$ $rarry=-79/3$

So, $x = - 16, y = - \frac{79}{3}$ $x=-16,y=-79/3$

How do you solve the following system?: 4x−3y=15,−6x+3y=174x-3y=15 , -6x+3y=17