How do you solve the system $2x-y= 7$ and $3x-4y= -7$ ?

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Answer 1 · 2018-05-19T18:41:31

$x = 7 and y = 7$

$2x - y = 7 - - - eqn1$

$3x - 4y = -7 - - - eqn2$

Multiply $eqn1$ by $3$ and $eqn2$ by $2$

$3 (2x - y = 7)$

$2 (3x - 4y = -7)$

$6x - 3y = 21 - - - eqn3$

$6x - 8y = -14 - - - eqn4$

Subtracting $eqn4$ from $eqn3$

$(6x - 6x) + (-3y - (-8y)) = 21 - (-14))$

$0 - 3y + 8y = 21 + 14$

$5y = 35$

$y = 35/5$

$y = 7$

Substiting the value of $y$ into $eqn1$

$2x - y = 7 - - - eqn1$

$2x - 7 = 7$

$2x = 7 + 7$

$2x = 14$

$x = 14/2$

$x = 7$

Therefore;

$x = 7 and y = 7$

How do you solve the system 2x-y= 72x-y= 7 and 3x-4y= -73x-4y= -7?