How to solve for y in $5 x - y = 33$ $5x-y=33$ and $7 x + y = 51$ $7x+y=51$ ?

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Answer 1 · 2015-04-08T01:41:44

$y = 5 x - 33$ $y=5x-33$ and $y = - 7 x + 51$ $y=-7x+51$ .

First Equation:

$5 x - y = 33$ $5x-y=33$

Subtract $5 x$ $5x$ from both sides.

$- y = 33 - 5 x$ $-y=33-5x$

Multiply both sides by $- 1$ $-1$ .

$y = - 33 + 5 x$ $y=-33+5x$

Rearrange.

$y = 5 x - 33$ $y=5x-33$

Second Equation:

$7 x + y = 51$ $7x+y=51$

Substitute the $y$ $y$ from the first equation

$7 x + (5 x - 33) = 51 \to 12 x - 33 = 51$ $7x+(5x-33)=51->12x-33=51$

Add $33$ $33$ to both sides, and then divide by $12$ $12$

$12 x = 84 \to x = 7$ $12x=84->x=7$

Now put this into the equation for $y$ $y$

$y = 5 x - 33 \to y = 5 \cdot 7 - 33 = 35 - 33 \to y = 2$ $y=5x-33->y=5*7-33=35-33->y=2$

Answer : $x = 7, y = 2$ $x=7,y=2$

How to solve for y in 5x−y=335x-y=33 and 7x+y=517x+y=51?