How do you solve the system of equations $3 x + y = 136$ $3x + y = 136$ and $x - 5 y = - 40$ $x - 5y = - 40$ ?

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Answer 1 · 2017-01-19T06:21:14

$3 x + y = 136$ $3x+y=136$
$x - 5 y = - 40$ $x-5y=-40$

$\Rightarrow x = 40$ $=> x=40$ and $y = 16$ $y=16$

This system

$3 x + y = 136$ $3x+y=136$
$x - 5 y = - 40$ $x-5y=-40$

$\to$ $to$ can also be written as

$((3,1),(1,-5))((x),(y))=((136),(-40))$ a matrix equation

and if a matrix is invertible it is true that

$Avecx=veccb => vecx=A^(-1)vecb$

To find the inverse $A^(-1)$

$A=((a,b),(c,d)) => A^(-1)=1/(det(A))((d,-b),(-c,a))$

In this case $a=3, b=1, c=1, d=-5$ . Therefore

$det(A)=ad-bc => det(A)=-15-1=-16$

following the formula for $A^(-1)$ we get

$=> A^(-1)=((5/16,1/16),(1/16,-3/16))$

$=> ((x),(y))=((5/16,1/16),(1/16,-3/16))((136),(-40))$

So we plug in and get

$=> x=((5)(136)-40)/16=640/16=40$

and

$y=(136+120)/16=256/16=16$

How do you solve the system of equations 3x + y = 1363x+y=1363x + y = 136 and x - 5y = - 40x−5y=−40x - 5y = - 40?