MATH SOLVE

3 months ago

Q:
# Solve the following system of equations using any method2x+6y+4z=−424x+3y+8z=−394x+3y+2z=3

Accepted Solution

A:

Answer:the values of x, y and z are: x=8, y = -5 and z = -7Step-by-step explanation:2x+6y+4z=−42 eq(1)
4x+3y+8z=−39 eq(2)
4x+3y+2z=3 eq(3)We would solve the above equations using elimination method.Subtracting eq(3) from eq(2)4x+3y+8z=−39
4x+3y+2z=3- - - -_____________0+0+6z = -42z = -42/6z = -7Multiplying eq(1) with 2 and subtracting with eq(2)4x + 12y +8z = -844x +3y +8z = -39- - - +_______________0+9y+0=-459y = -45y = -45/9y = -5Putting value of y and z in eq(1)2x + 6y +4z = -422x + 6(-5) +4(-7) = -422x -30 -28 = -422x -58 = -422x = -42 +582x = 16x = 16/2x= 8So, the values of x, y and z are: x=8, y = -5 and z = -7