Q 170.

Question

In the following exercises, solve the system of equations. 

3x-5y+4z=55x+2y+z=02x+3y-2z=3

Step-by-Step Solution

Verified
Answer

The solution is (2,-3,-4).

1Step 1. Given Information

We are given system of linear equations,

3x-5y+4z=5    -(1)5x+2y+z=0    -(2)2x+3y-2z=3    -(3)

2Step 2. Eliminating the same variable

Multiplying second equation by 4, we get

4(5x+2y+z)=4×020x+8y+4z=0   -(4)

Now, subtracting first and fourth equation, we get

20x-3x+8y+5y+4z-4z=0-517x+13y=-5   -(5)

Now, multiplying the second equation by 2, we get

2(5x+2y+z)=2×010x+4y+2z=0   -(6)

Now, adding sixth and third equation, we get

10x+2x+4y+3y+2z-2z=0+312x+7y=3   -(7)

3Step 3. Solving the new system of equation

Multiplying 13 in seventh equation and 7 in fifth equation, we get

13(12x+7y)=13(3)156x+91y=39   -(8)7(17x+13y)=7(-5)119x+91y=-35   -(9)

Now, subtracting eighth and ninth equation, we get

156x-119x+91y-91y=39-(-35)37x=74x=7437x=2

Now, putting the value of x in seventh equation, we get

12x+7y=312(2)+7y=324+7y=37y=3-24y=-217y=-3

Now, putting the value of x,y in second equation, we get

5x+2y+z=05(2)+2(-3)+z=010-6+z=0z=-4

4Step 4. Checking the solution

Putting the value of x,y,z in the equations, we get

3x-5y+4z=53(2)-5(-3)+4(-4)=56+15-16=55=55 x+2 y+z=05(2)+2(-3)-4=010-6-4=00=02 x+3 y-2 z=32(2)+3(-3)-2(-4)=34-9+8=33=3

Hence the solution is correct.