Question 631609
Given the following three equations to solve simultaneously:
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(A) 3x + 4y + z = 14
(B) .. + 2y + 7z = 24
(C) .. - 2y + 3z =  5
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Begin by adding equations B and C vertically. Note that the + 2y and the - 2y cancel each other out. The + 7z and the + 3z  add to give + 10z and on the other side of the equal sign the 24  and 5 add up to 29. So the result of adding these two equations is the new equation:
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10z = 29
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Solve this new equation for z by dividing both sides by 10 to get:
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z = 29/10 = 2.9
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Since you now know that z = 2.9, you can return to either equation B or equation C and, after substituting 2.9 for z, you can solve for y.  Go to equation B and substitute 2.9 for z to get:
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2y + 7*2.9 = 24
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Multiply out the 7 times 2.9 to get 20.3 and the equation becomes:
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2y + 20.3 = 24
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Subtract 20.3 from both sides of this equation and you are left with:
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2y = 3.7
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Solve for y by dividing both sides of this equation by 2 to get:
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y = 3.7/2 = 1.85
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Now you know that z = 2.9 and y = 1.85.  Equation A is the only equation that has a term containing x. So you next go to equation A and substitute 2.9 for z and 1.85 for y to get:
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3x + 4(1.85) + 2.9 = 14
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Multiply the 4 times 1.85 and the equation becomes:
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3x + 7.4 + 2.9 = 14
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Add the two numbers on the left side:
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3x + 10.3 = 14
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Subtract 10.3 from both sides:
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3x = 3.7
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Solve for x by dividing both sides by 3 to get:
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x = 3.7/3 = 1.23333333...
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So the values of x, y, and z that satisfy all three of the given equations simultaneously are x = 1.23333333..., y = 1.85, and z = 2.9
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Hope this helps you to understand the problem a little better. Check my work to ensure that it has no errors. You can also practice a little by using 2.9 for z and substituting that value into equation C. Then solve for y and see if you again get 1.85 for y just as you did by substituting 2.9 into equation B above.