How do you solve this system of equations: $3x + 5y = 19 and 4x - y = 10$ ?

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Answer 1 · 2017-11-28T14:10:49

$x=3 and y=2$

I will solve your system by substitution (you can also solve this system by elimination).

$4x−y=10$
$3x+5y=19$

Step: Solve $4x−y=10$ for $y$ :

$4x−y+−4x=10+−4x" "$ (add $-4x$ to both sides)

$−y=−4x+10$

Divide by $-1$

$y=4x−10$

Step: Substitute $4x−10$ for $y$ in $3x+5y=19$ :

$3x+5y=19$

$3x+5 (4x−10)=19$

$23x−50=19 " "$ (simplify both sides of the equation)

$23x−50+50=19+50 " "$ (add $50$ to both sides)

$23x=69$

Divide both sides by $23$

$x=3$

Step: Substitute $3$ for $x$ in $y=4x−10$ :

$y=4x−10$

$y=(4)(3)−10$

$y=2" "$ (simplify both sides of the equation)

How do you solve this system of equations: 3x + 5y = 19 and 4x - y = 103x + 5y = 19 and 4x - y = 10?