How do you solve the system $2 x + y = - 9$ $2x+y=-9$ and $3 x + y = 11$ $3x+y=11$ ?

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Answer 1 · 2015-05-24T09:25:30

Noticing the same term in $y$ $y$ in both equations, we can eliminate $y$ $y$ quickly as follows:

Subtract the first equation from the second to get:

$x = (3 x + y) - (2 x + y) = 11 - (- 9) = 20$ $x = (3x+y)-(2x+y) = 11 - (-9) = 20$

Substitute this value of $x$ $x$ into the first equation to get:

$- 9 = (2 \times 20) + y = 40 + y$ $-9 = (2xx20)+y = 40+y$

Subtract $40$ $40$ from both sides to get:

$y = - 9 - 40 = - 49$ $y = -9 -40 = -49$

As a quick check, substitute these values for $x$ $x$ and $y$ $y$ into the left hand side of the second equation:

$3 x + y = (3 \times 20) + (- 49) = 60 - 49 = 11$ $3x+y = (3xx20)+(-49) =60-49 = 11$

How do you solve the system 2x+y=-92x+y=−92x+y=-9 and 3x+y=113x+y=113x+y=11?