How do you solve the system of equations $4 x + 3 y = 3$ $4x + 3y = 3$ and $x - y = - 8$ $x - y = - 8$ ?

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How do you solve the system of equations $4 x + 3 y = 3$ $4x + 3y = 3$ and $x - y = - 8$ $x - y = - 8$ ?

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Answer 1 · 2017-05-02T23:24:45

$x = - 3 and y = 5$ $x=-3 and y=5$

Explanation:

$4 x + 3 y = 3$ $4x+3y=3$
$x - y = - 8$ $x-y=-8$
Multiply the second line by 3 to get rid of y.
$4 x + 3 y = 3$ $4x+3y = 3$
$3 x - 3 y = - 24$ $3x-3y=-24$
Add the two equations to get rid of y.
$7 x = - 21$ $7x=-21$
Now solve for x.
$\frac{7 x}{7} = \frac{- 21}{7}$ $(7x)/7=(-21)/7$
$x = - 3$ $x=-3$
Now sub x into one of the equations and solve for y.
$4 \cdot - 3 + 3 y = 3$ $4*-3+3y=3$
$3 y = 15$ $3y=15$
$\frac{3 y}{3} = \frac{15}{3}$ $(3y)/3=15/3$
$y = 5$ $y=5$
(-3,5)

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How do you solve the system of equations $4 x + 3 y = 3$ $4x + 3y = 3$ and $x - y = - 8$ $x - y = - 8$ ?

1 Answer

Explanation:

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