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# solve for (x) in this equation (3x^2)/(x^2-1)=(4x)/(x-1)

This looks like a quadratic equation, but the result are -2/3 and 2 which I dont know how to solve it.

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### x^2-1=0 - Equation Calculator - Symbolab

Equation Calculator Solve linear, ... 5x-6=3x-8; x^2-x-6=0; x^4-5x^2+4=0 \sqrt ... A quadratic equation is a second degree polynomial having the general form ax^2 ...

### How to solve for x in the equation [math]2x^2-1 = 2(x-1)^2 ...

... can also be written as [math]2x^2 - 4x + 2 + 4x -3 = 2 (x - 1)^2 + 4x - 3 ... I solve for x in the equation [math]2x^2-1 = 2 ... I solve the equation [math]x^3-3x+1=0

### How to solve |2x+1| < |x+2| - Quora

How do I solve for x in the equation [math]2x^2-1 = 2 ... 4x^2+4x+1 < x^2+4x+4 3x^2 < 3 x^2< 1 Here, x can be positive or negative both. So out solution will be:

### Solve For X: 4x^2+2x+2 = 0 - coolconversion.com

... Solve For X: 4x^2+2x+2 = 0 . Or how to solve the quadratic equation where ... to solve a equation like ax² ... Solve 4x^2-17x-3 = 0; Solve 3x^2+16x+1 = 0; Solve ...

### How do you solve 4x - 2( x - 4) = - 2+ 5x - 2? | Socratic

x=4. 4x-2(x-4)=-2+5x-2 is the problem. To solve it, ... out the -4 on the right side of the equation, giving us 8+4=3x-4 ... How do you solve #int x^-4*sqrt(x^4+1) ...

### How do you solve x/2 + x/3 - 1 = x/6 + 3? | Socratic

Now we move all of the fractions to one side of the equation and the whole numbers to the other: # ... How do you solve #4x ^ { 2} ... HOW do you solve #(x^2-1)/ (2 ...

### SOLUTIONS TO IMPLICIT DIFFERENTIATION PROBLEMS

SOLUTIONS TO IMPLICIT DIFFERENTIATION PROBLEMS ... (Equation 2) x 2 + y 2 = 1 ... (Now solve for y' .) 4x (x 2 +y 2) + 4y(x 2 +y 2) ...

### Mathway | Solve for x 2/(x+2)+1/5=6/(x+5)

Solve for x 2/(x+2)+1/5=6/(x+5) ... move it to the right side of the equation by subtracting from both sides. ... [x 2 1 2 π ∫ ⁡ x d x ...

### 3x+1=4x-2 - Get Easy Solution

Simple and best practice solution for 3x+1=4x-2 equation ... If it's not what You are looking for type in the equation solver your own equation and let us solve it ...

## Suggested Questions And Answer :

### how can i solve these equations?

how can i solve these equations? x + 3y – z = 2 x – 2y + 3z = 7 x + 2y – 5z = –21 We eliminate one of the unknowns (x, y or z, take your choice), leaving two unknowns. Then, we eliminate a second one, giving us an equation with only one of the unknowns. Solve for that and plug that value into one of the equations to solve for a second unknown. Finally plug both of those values into an equation to solve for the third unknown. It sounds complicated, but if you follow a logical sequence, the problem solves itself. 1) x + 3y – z = 2 2) x – 2y + 3z = 7 3) x + 2y – 5z = –21 If we subtract equation 3 from equation 2, we eliminate the x.    x – 2y + 3z =     7 -(x + 2y – 5z = –21) ------------------------     - 4y  + 8z =   28 4) -4y + 8z = 28 Subtract equation 1 from equation 2, eliminating the x again.    x – 2y + 3z = 7 -(x + 3y –   z = 2) ----------------------      - 5y +  4z = 5 5) -5y + 4z = 5 You now have two equations with only a y and a z. The easiest step now is to eliminate the z. Multiply equation 5 by 2. 2 * (-5y + 4z) = 5 * 2 6) -10y + 8z = 10 Subtract equation 6 from equation 4, eliminating the z.     -4y + 8z = 28 -(-10y + 8z = 10) ---------------------      6y      =   18 6y = 18 y = 3  <<<<<<<<<<<<<<<<<<<<< Plug that into equation 5 to solve for z. -5y + 4z = 5 -5(3) + 4z = 5 -15 + 4z = 5 4z = 20 z = 5  <<<<<<<<<<<<<<<<<<<<< Plug the values of y and z into equation 1 to solve for x. x + 3y – z = 2 x + 3(3) – 5 = 2 x + 9 - 5 = 2 x + 4 = 2 x = -2  <<<<<<<<<<<<<<<<<<<<< Always check the answers by plugging all three values into one of the original equations. Using all three would be even better. Equation 2: x – 2y + 3z = 7 (-2) – 2(3) + 3(5) = 7 -2 - 6 + 15 = 7 -8 + 15 = 7 7 = 7 Equation 3: x + 2y – 5z = –21 (-2) + 2(3) – 5(5) = –21 -2 + 6 - 25 = -21 6 - 27 = -21 -21 = -21 Answer: x = -2, y = 3, z = 5

### a sistem equation

a sistem equation 3x-5y-2z=-5 6x+2y+3z=5 4x-3y-z=8 x=? ;y=? ; z=? 1) 3x - 5y - 2z = -5 2) 6x + 2y + 3z = 5 3) 4x - 3y - z = 8 The way to solve systems like this is to sequentially eliminate each unknown so that there is only one left. That lets you solve for the value of that remaining unknown, and substitute that into one of the equations, then eliminate one of the other two unknowns. You then solve for the value of the second unknown. Now, you substitute both of those into one of the equations to solve for the third unknown. We'll use the first and second equation to eliminate the x term. Multiply equation 1 by 2. 2 * (3x - 5y - 2z) = -5 * 2 4) 6x - 10y - 4z = -10 Subtract equation 2 from equation 4.   6x - 10y -  4z = -10 -(6x +  2y + 3z =   5) -------------------------        - 12y - 7z = -15 5) -12y - 7z = -15 That gives us one equation with y and z. Use equations 2 and 3 to eliminate x again, giving us a second equation with y and z. We need to multiply equation 2 by 2, making 12 the co-efficient of x. Also, multiply equation 3 by 3, again making 12 the co-efficient of x. 2 * (6x + 2y + 3z) = 5 * 2 6) 12x + 4y + 6z = 10 3 * (4x - 3y - z) = 8 * 3 7) 12x - 9y - 3z = 24 Subtract equation 6 from equation 7.    12x -  9y - 3z = 24 -(12x + 4y + 6z = 10) --------------------------         - 13y -  9z = 14 8) -13y - 9z = 14 Using equations 5 and 8, we will eliminate the z term. Multiply equation 5 by 9. 9 * (-12y - 7z) = -15 * 9 9) -108y - 63z = -135 Multiply equation 8 by 7. 7 * (-13y - 9z) = 14 * 7 10) -91y - 63z = 98 Subtract equation 10 from equation 9.  -108y - 63z = -135  -(-91y - 63z =  98) ------------------------    -17y         = -233 y = 13.70588     <<<<<<<<<<<<<<<<<<<< Use that in equation 10 to solve for z. -91y - 63z = 98 -91(13.70588) - 63z = 98 -1247.23508 - 63z = 98 -63z = 98 + 1247.23508 -63z = 1345.23508 z = 1666/3367 / -63 z = -21.35293     <<<<<<<<<<<<<<<<<<<< Use equation 1 to solve for x. 3x - 5y - 2z = -5 3x - 5(13.70588) - 2(-21.35293) = -5 x = 6.94118     <<<<<<<<<<<<<<<<<<<< Use equation 2 to check the values. 6x + 2y + 3z = 5 6(6.94118) + 2(13.70588) + 3(-21.35293) = 5 41.64708 + 27.41176 - 64.05879 = 5 5.00005 = 5 Rounding errors account for the discrepancy. Use equation 3 to check the values. 4x - 3y - z = 8 4(6.94118) - 3(13.70588) - (-21.35293) = 8 27.76472 - 41.11764 + 21.35293 = 8 8.00001 = 8 Again, consider rounding errors. x = 6.94118; y = 13.70588; z = -21.35293

### Solve {3x-2y+2z=30, -x+3y-4z=-33, 2x-4y+3z=42}

Solve {3x-2y+2z=30, -x+3y-4z=-33, 2x-4y+3z=42} Please just solve the set provided above!!!! This will be a bit more involved than the systems with two unknowns, but the process is the same. The plan of attack is to use equations one and two to eliminate z. That will leave an equation with x and y. Then, use equations one and three to eliminate z again, leaving another equation with x and y. Those two equations will be used to eliminate x, leaving us with the value of y. I'll number equations I intend to use later so you can refer back to them. That's enough discussion for now. 1)  3x-2y+2z=30 2)  -x+3y-4z=-33 3)  2x-4y+3z=42 Equation one; multiply by 2 so the z term has 4 as the coefficient. 3x - 2y + 2z = 30 2 * (3x - 2y + 2z) = 30 * 2 4)  6x - 4y + 4z = 60 Add equation two to equation four:   6x - 4y + 4z =  60 +(-x + 3y - 4z = -33) ----------------------   5x - y       = 27 5)  5x - y = 27 Multiply equation one by 3. Watch the coefficient of z. 3 * (3x - 2y + 2z) = 30 * 3 6)  9x - 6y + 6z = 90 Multiply equation three by 2. Again, watch the coefficient of z. 2 * (2x - 4y + 3z) = 42 * 2 7)  4x - 8y + 6z = 84 Subtract equation seven from equation six.   9x - 6y + 6z = 90 -(4x - 8y + 6z = 84) ----------------------   5x + 2y      =  6 8)  5x + 2y = 6 Subtract equation eight from equation five. Both equations have 5 as the coefficient of x. We eliminate x this way.   5x -  y = 27 -(5x + 2y = 6) ---------------       -3y = 21 -3y = 21 y = -7  <<<<<<<<<<<<<<<<<<< At this point, I am confident that I followed the correct procedures to arrive at the value for y. Use that value to determine the value of x. ~~~~~~~~~~~~~~~ Plug y into equation five to find x. 5x - y = 27 5x - (-7) = 27 5x + 7 = 27 5x = 27 - 7 5x = 20 x = 4  <<<<<<<<<<<<<<<<<<< Plug y into equation eight, too. 5x + 2y = 6 5x + 2(-7) = 6 5x - 14 = 6 5x = 6 + 14 5x = 20 x = 4    same value for x, confidence high Proceed, solving for the value of z. ~~~~~~~~~~~~~~~ Plug both x and y into equation one. We will solve for z. Equation one: 3x - 2y + 2z = 30 3(4) - 2(-7) + 2z = 30 12 + 14 + 2z = 30 26 + 2z = 30 2z = 30 - 26 2x = 4 z = 2  <<<<<<<<<<<<<<<<<<< Continue using the original equations to check the values. Equation two: -x + 3y - 4z = -33 -(4) + 3(-7) - 4z = -33 -4 - 21 - 4z = -33 -25 - 4z = -33 -4z = -33 + 25 -4z = -8 z = 2   same value for z, looking good Equation three: 2x - 4y + 3z = 42 2(4) - 4(-7) + 3z = 42 8 + 28 + 3z = 42 36 + 3z = 42 3z = 42 - 36 3z = 6 z = 2  satisfied with the results We have performed several checks along the way, thus proving all three of the values. x = 4, y = -7 and z = 2

### 2x+y=9 x-2z=-3 2y+3z=15

2x+y=9 x-2z=-3 2y+3z=15 Forgot how to do This. Problems Solving systems of this type requires eliminating all but one unknown so you can solve for that unknown. Then, plug that value into one of the equations to solve for another of the unknowns. Finally, plug both of those values into an equation to solve for the last unknown. 1) 2x + y = 9 2) x - 2z = -3 3) 2y + 3z = 15 Multiply equation 2 by 2 so we can subtract it from equation 1. 2 * (x - 2z) = -3 * 2 4) 2x - 4z = -6   2x + y       =  9 -(2x     - 4z = -6) ---------------------         y + 4z = 15 5) y + 4z = 15 We now have two equations with y and z. Multiply equation 5 by 2... 2 * (y + 4z) = 15 * 2 6) 2y + 8z = 30 ...and subtract equation 6 from equation 3. Then, solve for z.   2y +  3z = 15 -(2y + 8z = 30) -------------------        - 5z = -15 -5z = -15 z = 3     <<<<<<<<<<<<<<<<<<<< Plug that into equation 3 to solve for y. 2y + 3z = 15 2y + 3(3) = 15 2y + 9 = 15 2y = 6 y = 3     <<<<<<<<<<<<<<<<<<<< Also, plug the z value into equation 2 to solve for x. x - 2z = -3 x - 2(3) = -3 x - 6 = -3 x = 3     <<<<<<<<<<<<<<<<<<<< Plug the values for x and y into equation 1 to verify that they are correct. 2x + y = 9 2(3) + 3 = 9 6 + 3 = 9 9 = 9 Usually, all three equations have all three unknowns, requiring more manipulation and more elimination, but the process is the same. For this problem, x = 3, y = 3 and z = 3.

### x - z = -3, y + z = 9, -2x + 3y +5z = 33

Problem: x - z = -3, y + z = 9, -2x + 3y +5z = 33 1) x - z = -3 2) y + z = 9 3) -2x + 3y + 5z = 33 Add equation 2 to equation 1.      x     - z = -3 +(    y + z =  9) ------------------   x + y     = 6 4) x + y = 6 Multiply equation 2 by 5. 5(y + z) = 9 * 5 5) 5y + 5z = 45 Subtract equation 3 from equation 5.             5y + 5z = 45 -(-2x + 3y + 5z = 33) --------------------------     2x + 2y        = 12 6) 2x + 2y = 12 Multiply equation 4 by 2. 2(x + y) = 6 * 2 7) 2x + 2y = 12 Subtract equation 7 from equation 6.    2x + 2y = 12 -(2x + 2y = 12) --------------------   0x      = 0      We are solving for x, not y x = 0    <<<<<<<<<<<<<<<<<<< Use equation 4 to solve for y. x + y = 6 0 + y = 6 y = 6    <<<<<<<<<<<<<<<<<<< Use equation 3 to solve for z. -2x + 3y + 5z = 33 -2(0) + 3(6) + 5z = 33 0 + 18 + 5z = 33 5z = 15 z = 3    <<<<<<<<<<<<<<<<<<< Check the values. 1) x - z = -3    0 - 3 = -3    -3 = -3 2) y + z = 9    6 + 3 = 9    9 = 9 3) -2x + 3y + 5z = 33    -2(0) + 3(6) + 5(3) = 33    0 + 18 + 15 = 33    33 = 33 Those numbers work..... ------------- What if we had solved for y instead of x here:    2x + 2y = 12 -(2x + 2y = 12) -------------------        0y = 0      We are solving for y, not x y = 0    <<<<<<<<<<<<<<<<<<< Use equation 4 to solve for x. x + y = 6 x + 0 = 6 x = 6    <<<<<<<<<<<<<<<<<<< Use equation 3 to solve for z. -2x + 3y + 5z = 33 -2(6) + 3(0) + 5z = 33 -12 + 0 + 5z = 33 5z = 45 z = 9    <<<<<<<<<<<<<<<<<<< Check the values. 1) x - z = -3    6 - 9 = -3    -3 = -3 2) y + z = 9    0 + 9 = 9    9 = 9 3) -2x + 3y + 5z = 33    -2(6) + 3(0) + 5(9) = 33    -12 + 0 + 45 = 33    33 = 33 Those numbers work, too. That complicates the solution. Answer 1: x = 0, y = 6, z = 3 Answer 2: x = 6, y = 0, z = 9

### x+2y+3z=14,x-y+2z =5, 2x + 2y +z= 11

x+2y+3z=14,x-y+2z =5, 2x + 2y +z= 11 1) x + 2y + 3z = 14 2) x - y + 2z = 5 3) 2x + 2y + z = 11 When there are three variables, it is easier to solve by elimination, but apparently this is to be solved by substitution. I'll number the equations as I go along, so you can see where I get each one I use later. Solve equation 1 for x. x + 2y + 3z = 14 4) x = -2y - 3z + 14 Substitute the value of x from equation 4 in place of the x in equation 2. x - y + 2z = 5 (-2y - 3z + 14) - y + 2z = 5 -2y - y - 3z + 2z + 14 = 5 -3y - z = 5 - 14 5) -3y - z = -9 Solve equation 5 for z. -3y - z = -9 -z = 3y - 9 6) z = -3y + 9 Now, it gets tricky. Using equation 4 again, substitute the value for x into equation 3. 2x + 2y + z = 11 2(-2y - 3z + 14) + 2y + z = 11 -4y - 6z + 28 + 2y + z = 11 -4y + 2y - 6z + z + 28 = 11 -2y - 5z + 28 = 11 -2y - 5z = 11 - 28 7) -2y - 5z = -17 Use the value of z from equation 6; substitute it into equation 7. -2y - 5z = -17 -2y - 5(-3y + 9) = -17 -2y + 15y - 45 = -17 13y = -17 + 45 13y = 28 y = 28/13 Substitute that into equation 6 to solve for z. z = -3y + 9 z = -3(28/13) + 9 z = -84/13 + 9 z = -84/13 + 117/13 z = 33/13 With values for y and z, we can solve for x. This calls for equation 1. x + 2y + 3z = 14 x + 2(28/13) + 3(33/13) = 14 x + 56/13 + 99/13 = 14 x + 155/13 = 14 x = 14 - 155/13 x = 182/13 - 155/13 x = 27/13 Check your work, using equation 2. x - y + 2z = 5 27/13 - 28/13 + 2(33/13) = 5 -1/13 + 66/13 = 5 65/13 = 5 5 = 5 Also, use equation 3. 2x + 2y + z = 11 2(27/13) + 2(28/13) + 33/13 = 11 54/13 + 56/13 + 33/13 = 11 143/13 = 11 11 = 11 The computed values are correct. Answer: x = 27/13, y = 28/13, z = 33/13

### 3x + 6y – 6z = 9 2x – 5y + 4z = 6 -x +16y + 14z = -3 what is the answer

3x + 6y – 6z = 9 2x – 5y + 4z = 6 -x +16y + 14z = -3 what is the answer how do you solve it and what are the answers ? 1)  3x + 6y – 6z = 9 2)  2x – 5y + 4z = 6 3)  -x +16y + 14z = -3 The first objective is to eliminate z so we can solve for x and y. It's a multi-step process, so follow along. Multiply equation one by 4. 4 * (3x + 6y – 6z) = 9 * 4 4)  12x + 24y - 24z = 36 Multiply equation 2 by 6. 6 * (2x – 5y + 4z) = 6 * 6 5)  12x - 30y + 24z = 36 Now, we have two equations with a "24z" term. Add the equations and the z drops out. Add equation five to equation four.    12x + 24y - 24z = 36 +(12x - 30y + 24z = 36) ----------------------------------    24x -  6y           = 72 6)  24x - 6y = 72 The same process applies to equations two and three. Multiply equation two by 7 this time. 7 * (2x – 5y + 4z) = 6 * 7 7)  14x - 35y + 28z = 42 Multiply equation three by 2. 2 * (-x + 16y + 14z) = -3 * 2 8)  -2x + 32y + 28z = -6 Subtract equation eight from equation seven.   14x - 35y + 28z = 42 -(-2x + 32y + 28z = -6) ---------------------------------   16x - 67y          = 48 9)  16x - 67y = 48 Looking at equations six and nine, it would be simpler to eliminate the x. The multipliers are smaller. Multiply equation six by 2. 2 * (24x - 6y) = 72 * 2 10)  48x - 12y = 144 Multiply equation nine by 3. 3 * (16x - 67y) = 48 * 3 11)  48x - 201y = 144 Subtract equation eleven from equation 10.   48x -   12y = 144 -(48x - 201y = 144) ---------------------------          189y  =   0 189y = 0 y = 0  <<<<<<<<<<<<<<<<<<<<<<<<<<<<<< Substitute that value into equations six and nine to solve for x and verify. Six: 24x - 6y = 72 24x - 6(0) = 72 24x - 0 = 72 24x = 72 x = 3  <<<<<<<<<<<<<<<<<<<<<<<<<<<<<< Nine: 16x - 67y = 48 16x - 67(0) = 48 16x - 0 = 48 16x = 48 x = 3    same answer for x To solve for z,substitute the x and y values into the three original equations. One: 3x + 6y – 6z = 9 3(3) + 6(0) – 6z = 9 9 + 0 - 6z = 9 9 - 6z = 9 -6z = 9 - 9 -6z = 0 z = 0  <<<<<<<<<<<<<<<<<<<<<<<<<<<<<< Two: 2x – 5y + 4z = 6 2(3) – 5(0) + 4z = 6 6 - 0 + 4z = 6 6 + 4z = 6 4z = 6 - 6 4z = 0 z = 0     same answer Three: -x +16y + 14z = -3 -(3) +16(0) + 14z = -3 -3 + 0 + 14z = -3 -3 + 14z = -3 14z = -3 + 3 14z = 0 z = 0     once again, same answer x = 3, y = 0, z = 0

### solve following system of equations

Problem: solve following system of equations solve the following system of equations: x-2y+z=6 , 2x+y-3z=-3, x-3y+3z=10 1) x - 2y + z = 6 2) 2x + y - 3z = -3 3) x - 3y + 3z = 10 Multiply equation 1 by 3. 3(x - 2y + z) = 6 * 3 4) 3x - 6y + 3z = 18 Add equation 2 to equation 4.   3x - 6y + 3z = 18 +(2x +  y - 3z = -3) -----------------------   5x - 5y      = 15 6) 5x - 5y = 15 Add equation 3 to equation 2.   2x +  y - 3z = -3 +( x - 3y + 3z = 10) ----------------------   3x - 2y      = 7 7) 3x - 2y = 7 Multiply equation 6 by 2. 2(5x - 5y) = 15 * 2 8)10x - 10y = 30 Multiply equation 7 by 5. 5(3x - 2y) = 7 * 5 9) 15x - 10y = 35 Subtract equation 9 from equation 8.   10x - 10y = 30 -(15x - 10y = 35) -------------------   -5x       = -5 -5x = -5 -5x/-5 = -5/-5 x = 1 Use equation 6 to solve for y. 5x - 5y = 15 5(1) - 5y = 15 5 - 5y = 15 5 - 5y - 5 = 15 - 5 -5y = 10 -5y/-5 = 10/-5 y = -2 Use equation 3 to solve for z. x - 3y + 3z = 10 1 - 3(-2) + 3z = 10 1 + 6 + 3z = 10 7 + 3z = 10 3z = 3 3z/3 = 3/3 z = 1 Answer: x = 1, y = -2, z = 1

### using the elimination method solve 6=6x+6y and -14y=18x+14

using the elimination method solve 6=6x+6y and -14y=18x+14 help me solve step by step 1) 6 = 6x + 6y 2) -14y = 18x + 14 Re-write equation 1: 3) 6x + 6y = 6 It is standard to have the unknowns of the left side of the equation. Now, for equation 2. -14y = 18x + 14 4) -18x - 14y = 14 Multiply equation 3 by 3. 3 * (6x + 6y) = 6 * 3 5) 18x + 18y = 18 Add equation 4 to equation 5.    18x + 18y = 18 +(-18x - 14y = 14) -------------------           4y = 32 You have just eliminated the x. Solve for y. 4y = 32 y = 8 You can used either equation 1 or equation 2 to solve for x. 1) 6 = 6x + 6y = 6    6x + 6(8) = 6    6x + 48 = 6    6x = 6 - 48    6x = -42    x = -7 Check it with equation 2. 2) -14y = 18x + 14    -14(8) = 18(-7) + 14    -112 = -126 + 14    -112 = -112 You have x = -7 and y = 8

### solve three way equation with z. 1:3x-2y+2z=30 2:-x+3y-4z=-33 3:2x-4y+3z=42

Solve {3x-2y+2z=30, -x+3y-4z=-33, 2x-4y+3z=42} Please just solve the set provided above!!!! 1)  3x-2y+2z=30 2)  -x+3y-4z=-33 3)  2x-4y+3z=42 Equation one; multiply by 2 so the z term has 4 as the coefficient. 3x - 2y + 2z = 30 2 * (3x - 2y + 2z) = 30 * 2 4)  6x - 4y + 4z = 60 Add equation two to equation four:   6x - 4y + 4z =  60 +(-x + 3y - 4z = -33) ----------------------   5x - y       = 27 5)  5x - y = 27 Multiply equation one by 3. Watch the coefficient of z. 3 * (3x - 2y + 2z) = 30 * 3 6)  9x - 6y + 6z = 90 Multiply equation three by 2. Again, watch the coefficient of z. 2 * (2x - 4y + 3z) = 42 * 2 7)  4x - 8y + 6z = 84 Subtract equation seven from equation six.   9x - 6y + 6z = 90 -(4x - 8y + 6z = 84) ----------------------   5x + 2y      =  6 8)  5x + 2y = 6 Subtract equation eight from equation five. Both equations have 5 as the coefficient of x. We eliminate x this way.   5x -  y = 27 -(5x + 2y = 6) ---------------       -3y = 21 -3y = 21 y = -7  <<<<<<<<<<<<<<<<<<< ~~~~~~~~~~~~~~~ Plug y into equation five to find x. 5x - y = 27 5x - (-7) = 27 5x + 7 = 27 5x = 27 - 7 5x = 20 x = 4  <<<<<<<<<<<<<<<<<<< Plug y into equation eight, too. 5x + 2y = 6 5x + 2(-7) = 6 5x - 14 = 6 5x = 6 + 14 5x = 20 x = 4    same value for x Proceed, solving for the value of z. ~~~~~~~~~~~~~~~ Plug both x and y into equation one. We will solve for z. Equation one: 3x - 2y + 2z = 30 3(4) - 2(-7) + 2z = 30 12 + 14 + 2z = 30 26 + 2z = 30 2z = 30 - 26 2x = 4 z = 2  <<<<<<<<<<<<<<<<<<< Continue using the original equations to check the values. Equation two: -x + 3y - 4z = -33 -(4) + 3(-7) - 4z = -33 -4 - 21 - 4z = -33 -25 - 4z = -33 -4z = -33 + 25 -4z = -8 z = 2   same value for z Equation three: 2x - 4y + 3z = 42 2(4) - 4(-7) + 3z = 42 8 + 28 + 3z = 42 36 + 3z = 42 3z = 42 - 36 3z = 6 z = 2  satisfied with the results   x = 4, y = -7 and z = 2