How do you solve $x / 2 - y / 3 = 5/6$ and $x / 5 - y / 4 = 29 / 10$ using substitution?

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Answer 1 · 2016-07-16T17:29:10

$x= -13, y = -22$

Before we can even think of solving, let's change the equations into a more user-friendly form!.

Multiply each by the LCM of the denominators to get rid of the fractions.

$(color(magenta)6 xx x) / 2 - (color(magenta)6 xx y) / 3 = (color(magenta)6 xx 5)/6 " and " "x / 5 - y / 4 = 29 / 10$

$3x - 2y = 5 " and "(color(blue)20 xx x) / 5 - (color(blue)20 xx y) / 4 = (color(blue)20 xx 29)/10$

$3x - 2y = 5 " and " 4x -5y = 58$

Make one of the variables in one equation the subject.

$color(magenta)(x = (5+2y)/3 )" or " y = (3x-5)/2$

(one is not really easier than the other to substitute.)

$4 (color(magenta)((5+2y)/3 )) -5y = 58$

$((20+8y))/3 - 5y = 58" " xx3$

$20+8y -15y = 174$

$20-174 = 7y$

$-154 = 7y$

$y =-22 " now use this value to find x"$

$x = (5+2y)/3 rArr " " x = (5+2(-22))/3$

$x = -13$

Check these values in the second equation to verify.

How do you solve x / 2 - y / 3 = 5/6x / 2 - y / 3 = 5/6 and x / 5 - y / 4 = 29 / 10x / 5 - y / 4 = 29 / 10 using substitution?