How do you solve this system of equations: $x + 3 y = - 4 and 5 x + 4 y = 13$ $x+ 3y = - 4 and 5x + 4y = 13$ ?

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Answer 1 · 2017-11-11T06:58:45

Solve for either x or y and substitute the value of the variable in the other equation to solve for which variable is left.

$x + 3 y = - 4$ $x+3y = -4$

$5 x + 4 y = 13$ $5x +4y = 13$

I will Solve for $x$ $x$ first because I see smaller numbers that are easier to deal with.

$x + 3 y = - 4$ $x + 3y = - 4$

subtract $3 y$ $3y$ from both sides to get $x$ $x$ isolated

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

take this solution and plug it in for $x$ $x$ in the other equation

$5 (- 4 - 3 y) + 4 y = 13$ $5(-4 - 3y) + 4y = 13$

distributive property

$- 20 - 15 y + 4 y = 13$ $-20 -15y + 4y = 13$

add like terms

$- 20 - 11 y = 13$ $-20-11y = 13$

add $20$ $20$ to both sides to isolate y

$- 11 y = 33$ $-11y = 33$

divide both sides by a form of one to Isolate variable further

$\frac{- 11 y}{- 11} = \frac{33}{- 11}$ $(-11y)/-11 = 33/-11$

$y = - 3$ $y = -3$

for $x = - 4 - 3 y$ $x=-4-3y$ substitute $y$ $y$ to get $x$ $x$

$- 4 - 3 (- 3) = 5$ $-4-3(-3) = 5$

the solutions to this system of equations are

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

How do you solve this system of equations: x+3y=−4and5x+4y=13x+ 3y = - 4 and 5x + 4y = 13?