How do you solve the system of equations $2x - 5y = 3$ and $3x - 6y = 9$ ?

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Answer 1 · 2018-04-07T01:40:49

$x = 9$
$y = 3$

To solve this system of equations, we will first manipulate the first equation so that $y$ is alone on the left-hand side of the equation.

$2x - 5y = 3$ ,
$-5y = 3 - 2x$ ,
$5y = -3 + 2x$ ,
$5y = 2x - 3$ ,
$y = 2/5 x - 3/5$ .

We now plug this value of $y$ into the second equation.

$3x - 6y = 9$ ,
$3x - 6(2/5 x - 3/5) = 9$ ,
$3x - 12/5 x + 18/5 = 9$ .

Move all the $x$ 's to one side of the equation and the constants to the other.

$3x - 12/5 x + 18 /5 = 9$ ,
$3x - 12/5 x = 9 - 18/5$ .

Multiplying both sides of the equation by $5$ makes it easier to manipulate.

$5(3x - 12/5 x) = 5(9 - 18 / 5)$ ,
$15x - 12x = 45 - 18$ ,
$3x = 27$ ,
$x = 9$ .

Now plug this $x$ back into the first equation.

$2x - 5y = 3$ ,
$2(9) - 5y = 3$ ,
$18 - 5y = 3$ ,
$18 = 3 + 5y$ ,
$15 = 5y$ ,
$3 = y$ .

Thus, we have $x=9$ and $y=3$ .

How do you solve the system of equations 2x - 5y = 32x - 5y = 3 and 3x - 6y = 93x - 6y = 9?