SOLUTION: Solve for X,Y and Z X+10Y+Z=52 5X+Y+4Z=15 X+2y-3Z=12

Question 202802: Solve for X,Y and Z
X+10Y+Z=52
5X+Y+4Z=15
X+2y-3Z=12

Answer by PRMath(133) (Show Source): You can put this solution on YOUR website!
Solve for X,Y and Z
X+10Y+Z=52
5X+Y+4Z=15
X+2y-3Z=12

First let's choose ANY two equations and eliminate ONE variable. I'd suggest the first and 3rd equations..........

X+10Y+Z=52
X+2y-3Z=12 (Hm... what if we multiplied this equation by -1?)

X+10Y+Z=52
-X-2y+3Z=-12 (Do you see how we multiplied this whole equation by -1?)
_____________ Now let's add these equations
8y + 4z = 40 (Do you see how the "X" variables were eliminated?)

Now let's take another two equations and eliminate the SAME variable. I'd suggest the 2nd and 3rd variable. Let's eliminate for "X" because that is what we eliminated above......

5X + Y + 4Z = 15
X +2y -3Z = 12 (Hmm.. what if we multiplied this whole equation by -5)?

If we multiply the 2nd equation by -5, the two equations will then be....

5X + Y + 4Z = 15
-5x- 10y+15z = -60
__________________ Now let's add these.........
-9y + 19z = -45

OHhhh.. now we have two equations that have two variables. Let's work with them...........

8y + 4z = 40 (Hm... what if we multiply this whole line by 9?)
-9y + 19z = -45 (What if we multiply this whole line by 8?)

If we did the multiplications I suggest above, we'd get:

72y + 36z = 360
-72y + 152z = -360
_____________________ Let's add these equations....
0y + 188z = 0
z = 0 (divided both sides by 188)

Now we know that z = 0. Let's plug that number into one of our 2 variable equations......

8y + 4z = 40
8y + 4(0) = 40
8y = 40
y = 5 (divided both sides by 8)

Now we know that z = 0 and y = 5. Let's plug those numbers into one of our 3 variable equations..............

X + 2y -3Z = 12 (one of our original 3 variable equations)
x + 2(5) -3(0)= 12 (plugged in our numbers for variables)
x + 10 = 12 (multiplied 2 times 5 and -3 times 0)
x = 12 - 10 (subtracted 10 from both sides)
x = 2

We now know that x = 2, y = 5 and z = 0. If we have worked correctly, we can plug these in for the variables on our original three equations. Let's check.........

Original equation 1

X+10Y+Z=52 (original equation)
2 + 10(5) + 0 = 52
2 + 50 + 0 = 52
52 = 52
Yay! The 1st equation works out correctly.

Second original equation

5X+Y+4Z=15
5(2) + 5 + 4(0)= 15
10 + 5 + 0 = 15
15 = 15
Yay again! This 2nd equation works out correctly. Let's go for the 3rd!

Third original equation

X+2y-3Z=12
2 + 2(5) - 3(0) = 12
2 + 10 - 0 = 12
12 = 12
Yayyyyy for the third right answer. So our x, y, and z values are correct.

I hope this helps. :-)
RELATED QUESTIONS
Please solve these three system equation for X and Y and Z. Equation1. X+10Y+Z=52... (answered by JBarnum)

X+10Y+Z=52 5X+Y+4Z=15 X+2y-3Z=12 (answered by Edwin McCravy)

Solve for x,y in the following system of three x + 10y + z = 52 5x + y + 4z =15 x + 2y (answered by venugopalramana)

Solve for X, Y, and Z in the following systems of three equations: Can you show me the... (answered by Edwin McCravy)

Need help with the problem below.Thanks in advance. Systems of Equations Solve the... (answered by Fombitz)

1) x+y=9 x+2y=13 x= y= 2) 3x+7y= 24 9x+5y=24 x= y= 3) 4x+y=17 2x+3y+21 x= (answered by Fombitz)

Systems of Equations 1. Solve the following problems for X and Y i these two system... (answered by Alan3354)

1.Solve the following problems fo X and Y in these two system equation, and for X and Y,... (answered by solver91311)

Solve for x, y, and z variables of this systems of equations problem: 3x + 2y + z = 7, 5x (answered by ewatrrr)