How do you solve this 2x + 5y + 2z =6 5x -7y =-29 z=1?

Question

How do you solve this 2x + 5y + 2z =6 5x -7y =-29 z=1?

Answer 1

This may look like a lot to read, but please don't ignore it! I have given a step-by-step instruction guide on how to go about solving this problem!

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First of all, get rid of the z-term in equation 1 by plugging in the value of z. Isolate the 2z (which becomes 2 after plugging in z) by subtracting it from both sides.Your resulting 2 equations are...

2x + 5y = 4

5x - 7y = -29

Now that that's out of the way, lets move on to the meat of this problem...

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First, solve one of the equations (it doesn't matter which) for either x or y (again, either one is fine). I chose to solve the first equation for x and got x = -5/2y + 2. Next, plug this value into the other equation for the variable you have solved for (in this case, I will plug in -5/2y + 2 for x). Your resulting equation is as so:

5(-5/2y + 2) - 7y = -29

As you can see, I have substituted the x for -5/2y + 2. I can do this because we figured out when isolating x that x = -5/2y + 2. Therefore, the value of the equation isn't changed when making this substitution in other equations. Moving on...

Our next step is to solve this equation,

5(-5/2y + 2) - 7y = -29, for y. It will take quite a few steps, but eventually y will be isolated to 2. You now have your value for y: y=2. Now all you have to do is substitute this value in for y in your other equation, 2x + 5y = 4. Your equation, once you have plugged in the y value, will appear as such:

2x + 5(2) = 6

Solve this equation for x, and you will have solved the problem! (I am leaving this last step for you to do)

PS - Remember, as I mentioned in the beginning of this answer, you did not have to follow is particular method. You could have substituted a different value, and you could have also begun with the other equation.

PPS - glurpy: you did your math right and all, but you accidentally worte x=-1 instead of x=-3 for your final answer.

Answer 2

Another way to solve this is through the elimination method.

First, simplify your first equation:

2x + 5y + 2(1) = 6

2x + 5y + 2 = 6

2x + 5y + 2 - 2 = 6 - 2

2x + 5y = 4

Now you have the following equations:

2x + 5y = 4

5x - 7y = -29

To eliminate, you need to have the same number of x's or y's. Let's go for x's. The LCM for 2 and 5 is 10, so we'll multiply each equation so we have 10x:

5(2x + 5y = 4)

2(5x - 7y = -29)

We rewrite our equations:

10x + 25y = 20

10x - 14y = -58

Since we want to get rid of the x's (eliminate them), we'll subtract the second line from the first line and get this:

39y = 78

y = 2

We then plug y into any of the original equations to figure out x:

5x - 7(2) = -29

5x - 14 = -29

5x - 14 + 14 = -29 + 14

5x = -15

x = -3

We can check with the final equation:

2(-3) + 5(2) + 2(1) = 6

-6 + 10 + 2 = 6

6 = 6

Therefore:

x=-1

y=2

z=1