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

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Answer 1 · 2017-11-28T14:10:49

$x = 3 and y = 2$ $x=3 and y=2$

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

$4 x - y = 10$ $4x−y=10$
$3 x + 5 y = 19$ $3x+5y=19$

Step: Solve $4 x - y = 10$ $4x−y=10$ for $y$ $y$ :

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

$- y = - 4 x + 10$ $−y=−4x+10$

Divide by $- 1$ $-1$

$y = 4 x - 10$ $y=4x−10$

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

$3 x + 5 y = 19$ $3x+5y=19$

$3 x + 5 (4 x - 10) = 19$ $3x+5 (4x−10)=19$

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

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

$23 x = 69$ $23x=69$

Divide both sides by $23$ $23$

$x = 3$ $x=3$

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

$y = 4 x - 10$ $y=4x−10$

$y = (4) (3) - 10$ $y=(4)(3)−10$

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

How do you solve this system of equations: 3x + 5y = 19 and 4x - y = 103x+5y=19and4x−y=103x + 5y = 19 and 4x - y = 10?