How to solve this three variable systems with elimination. 12x+24y-8z=-24, 4x+2y+8z=38,-5x-2y+8z=62

Answer 1

This is a series of three equations that you are being asked to solve for the three unknowns—x, y, and z. This happens to be set up quite conveniently. If you add the first equation to the second, the z term will drop out and you will have one equation in two unknowns, x and y. Set that equation aside. 16x + 26y = 14

Next, add the first equation to the third equation. Again, the z term will drop out, and you will have an equations with only two unknowns, x and y. 7x + 22y = 38

Now just consider those two equations with x and y that created. I'm going to simplify the problem by dividing one equation by 2, since each of the terms is evenly divisible by two. This gives us 8x + 13y = 7, and will simplify our lives. Now we need to solve the two equation below by, as required, elimination.

8x + 13y = 7

7x + 22y = 38

Since the x coefficients (8 and 7) are smaller than the y coefficients (13 and 22), I'm going to work our math magic on the smaller coefficients. This means, I'm going to eliminate the x terms. I do this by multiplying each equation by a number that will cause the new coefficients of x to be opposite, one positive and one negative.

–7(8x + 13y) = 7 (–7)

8(7x + 22y) = 38 (8)

Doing the distribution over addition on the left and the multiplication on the right gives us:

–56x – 91y = –49

56x + 176y = 304

Now, add those two equations together, keeping the right side separated from the right side by an equal sign and combining like terms to get:

85y = 255

Now divide each side by 85 to find out what one y equals.

85y 255

------ = ------- and so y= 3

85 85

I'm hoping that it is okay to use substitution now.

I would go back to 8x + 13y = 7 and replace the y with 3, since they are equal and you'll be able to solve for x.

8x + 13(3) = 7

8x + 39 = 7

8x = 7 –39

8x = –32

x = –4

So x = –4 and y = 3, so you can substitute those into one of the three original equations to determine what z equals. You can take whatever of the three you wish. I am going to take the second for two reasons: (1) it has no negatives; (2) I can divide through by 2 at the start to make for simpler math.

4x + 2y + 8z = 38

Divide everything by two to get:

2x + y + 4z = 19

Substitute –4 for the x and 3 for the y to get:

2(–4) + 3 + 4z = 19

–8 + 3 + 4z = 19

–5 + 4z = 19

4z = 24

z = 6

You can give your answer as (–4, 3, 6) or as x = –4, y = 3, z = 6.

I always like to check my answer by substituting those values into each of the three equations to make sure that make true statements, but I will leave that to you.

Answer 2

The goal here is to add multiples of rows to each other until you only have one variable in each row.

It's easiest to start out writing our equations stacked like this.

12x+24y-8z=-24

4x+2y+8z=38

-5x-2y+8z=62

Here, I'm adding row 2 to row 3 in order to get rid of the y term in row 3.

12x+24y-8z=-24

4x+2y+8z=38

-x+0+16z=100

Here, I'm adding 4 times row 3 to row 2 to get rid of the x term in row 2.

12x+24y-8z=-24

0+2y+72z=438

-x+0+16z=100

Here, I'm adding 12 times row 3 to row 1 to get rid of the x term in row 1.

0+24y+184z=1,176

0+2y+72z=438

-x+0+16z=100

Here, I'm subtracting 12 times row 2 from row 1 to get rid of the y term in row 1.

0+0-680z=-4,080

0+2y+72z=438

-x+0+16z=100

Here, I'm dividing row 1 by -680 to get rid of it. In the previous step, -680 would be referred to as the leading coefficient of row 1.

0+0+z=6

0+2y+72z=438

-x+0+16z=100

Here, I'm dividing row 2 by 2 to get rid of the leading coefficient of row 2.

0+0+z=6

0+y+36z=219

-x+0+16z=100

Here, I'm multiplying row 3 by -1 to get rid of the leading coefficient of row 3.

0+0+z=6

0+y+36z=219

x+0-16z=-100

Here, I'm subtracting 36 times row 1 from row 2 to get rid of the z term in row 2.

0+0+z=6

0+y+0=3

x+0-16z=-100

Here, I'm adding 16 times row 1 to row 3 to get rid of the z term in row 3.

0+0+z=6

0+y+0=3

x+0+0=-4

And now we have our solution.

z=6

y=3

x=-4