# 3x+5y+4=13 5x+2y+3z=-9 6x+3y+4z=-8

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3x+5y + 4z = 13.............(1)

5x+2y + 3z = -9............(2)

6x + 3y + 4z = -8................(3)

We will use elimination and substitution method to solve/

First we will subtract (1) from (3).

==> 3x -2y = -21 .............(4)

Now we will multiply (1) by -3 and multiply (2) by 4 and add.

==> -9x -15y -12z = -39

==> 20x + 8y +12z = -36

==> 11x - 7y = -75..................(5)

Now we will solve for (4) and (5).

== -7(4) + 2(5)

==> -21x +14y = 147

==> 22x -14y = -150

==> x = -3

==> Now we will use equation (4) to find the value of y.

==> 3x-2y= -21

==> 3(-3) -2y = -21

==> -9 -2y= -21

==> 2y= -9+21= 12

==> y= 6

Now we will substitute x= -3 and y= 6 into (1) to find z.

===> 3x + 5y + 4z = 13

==> 3(-3) + 5(6)+ 4z = 13

==> -9 + 30 + 4z = 13

==> 4z = 13+9 -30 = -8

==> z = -2

**Then the solution to the system is (x, y, z)= (-3,6, -2)**