SOLUTION: Solving a system of 3 equations by elimination. 5x+2y=4 3x+4y+2z=6 7x+3y+4z=29 I took 3x=4y+2z and multiplied it by 7 getting 21x+28y+14z=42 then did 7x+3y+4z=29 and multiplied

Algebra ->  College  -> Linear Algebra -> SOLUTION: Solving a system of 3 equations by elimination. 5x+2y=4 3x+4y+2z=6 7x+3y+4z=29 I took 3x=4y+2z and multiplied it by 7 getting 21x+28y+14z=42 then did 7x+3y+4z=29 and multiplied      Log On


   



Question 520397: Solving a system of 3 equations by elimination. 5x+2y=4 3x+4y+2z=6 7x+3y+4z=29
I took 3x=4y+2z and multiplied it by 7 getting 21x+28y+14z=42 then did 7x+3y+4z=29 and multiplied by 3 to get 21x+9y+12z=87 but i don't know what to do next.

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Given three equations with three unknowns as follows:
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Equation A: 5x+%2B+2y+=+4
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Equation B: 3x+%2B+4y+%2B+2z+=+6
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Equation C: 7x+%2B+3y+%2B+4z+=+29
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What you did so far was correct. But with a little planning you could save some work. Here's the planning part. First, the goal is to start eliminating variables so that you keep ending up with equations that have less variables to solve for.
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Notice that Equation A already has a missing variable. There is no term having a "z" in it and therefore, we already have some of the work done for us. We need a second equation missing the "z" term. How can we do that??? Equations B and C both have terms involving "z". How could we combine Equations B and C such that we caused the resulting equation to have no term with "z" in it?
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Suppose we doubled every term (both left and right side). If we multiplied Equation B by 2 we would make Equation B become:
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6x+%2B+8y+%2B+4z+=+12
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Notice that by doubling Equation B we made the "z" term become "4z" and this is the same value as the "z" term in Equation C. Now if we subtract Equation C from our new Equation B we get:
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6x+%2B+8y+%2B+4z+=+12
-7x+-+3y+-+4z+=+-29
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Notice that I multiplied all the terms in Equation C by minus 1. Now instead of subtracting I just vertically add the terms in the bottom equation to the terms in the top equation. Look specifically at what happens in the "z" column. The +4z of the top equation is canceled by the -4z from the bottom equation. Therefore, a "z" term will not appear. In the first column, we have +6x on top and -7x below it. When we add those two, the result is -x. In the "y" column, we have +8y on top and -3y below it. When we add those two the result is +5y. And as we said previously, we multiplied B by 2 because it made the "z" term become +4z and subtracting the 4z term in Equation C will make the "z" terms disappear. Finally, on the right side we add +12 and -29 (this is the same as subtracting +29 from +12) and we get an answer of -17. So when we subtract Equation C from our "new" doubled Equation B we get:
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-x+%2B+5y+=+-17
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Now we have another equation with no "z" term. Recall that Equation A also had no "z" term. Therefore, the two equations without "z" terms are Equation A and this combined equation that resulted from combining Equations B and C such that we eliminated the "z" term. Let's write those two equations so they appear in columns:
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-x+%2B+5y+=+-17 (Combined Equations B and C}
5x+%2B+2y+=+4 (Original equation A)
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Notice that if we multiply the top equation by 5 we will make the "x" terms in these two equations equal but of opposite signs:
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-5x+%2B+25y+=+-85 (multiplied by 5)
5x+%2B+2y+=+4
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Since the "x" terms in the two equations are equal but of opposite signs, we can get rid of these "x" terms by just adding the equations in vertical columns as follows:
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-5x+%2B+25y+=+-85
5x+%2B+2y+=+4
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The -5x and +5x terms cancel each other and we are left with:
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27y+=+-81
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We can solve for y by dividing both sides of this resulting equation by 27 and we have:
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27y%2F27+=+-81%2F27
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and this reduces to:
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y+=+-3
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Now that we know that y is -3, we can return to one of the equations that contained only the variables "x" and "y", substitute -3 for y, and solve for x.
Let's return to Equation A because it has only those two variables. And substitute into Equation A -3 for y. When we do that we have:
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5x+%2B+2%2A%28-3%29+=+4
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Multiply out the left side term to get:
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5x+-6+=+4
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Add 6 to both sides and to eliminate the -6 and you have:
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5x+=+10
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Solve for x by dividing both sides by 5:
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x+=+2
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Now that we know x is +2 and y is -3, we can return to the original equations (either Equation B or Equation C), substitute the known values for x and y and solve for z. Let's use Equation B with substitutions to get:
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3%2A%282%29+%2B+4%2A%28-3%29+%2B+2z+=+6
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Multiply out the first two terms on the left side:
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6+-12+%2B2z+=+6
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Combine the 6 and -12 and you have:
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-6+%2B+2z+=+6
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Get rid of the -6 by adding +6 to both sides:
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2z+=+12
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Solve for z by dividing both sides by 2 and you have:
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z+=+6
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That's the last piece of the puzzle. The solution to this problem is x = 2, y = -3, and z = 6.
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You can check this out by going back to the original equations and substituting these values into each of the equations to makes sure that they balance with the left side equaling the right side.
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Hope this solution process gives you some ideas you can use to solve equations by eliminating variables. Don't lose sight of the plan. Use multiplication to get terms in two equations equal but of opposite sign and then add to eliminate those terms. You just have to keep working down through the equations until only one variable is left. Once you solve it, reverse the process and start working backward to solve for each of the other variables.
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