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Solve the following system of equations algebraically using linear combination and substitution. You may use a calculator to check your answer, all work in solving must be written

x+y-3z=8
2x-3y+z=-6
3x+4y-2z=20

User ManiAm
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Answer:To solve the system of equations using linear combination and substitution, we'll follow these steps:

Step 1: Choose two equations and eliminate one variable.

Let's choose the first and second equations. We can eliminate the variable z by multiplying the first equation by 3 and the second equation by -1, and then adding them together.

3(x+y-3z) = 3(8) -> 3x+3y-9z=24

-1(2x-3y+z) = -1(-6) -> -2x+3y-z=6

Adding these two equations gives us:

(3x+3y-9z) + (-2x+3y-z) = 24+6

Simplifying, we get:

x+6y-10z = 30 (Equation 3)

Step 2: Choose two different equations and eliminate the same variable.

Now, let's choose the first and third equations. We can eliminate the variable x by multiplying the first equation by -3 and the third equation by 1, and then adding them together.

-3(x+y-3z) = -3(8) -> -3x-3y+9z=-24

1(3x+4y-2z) = 1(20) -> 3x+4y-2z=20

Adding these two equations gives us:

(-3x-3y+9z) + (3x+4y-2z) = -24+20

Simplifying, we get:

y+7z = -4 (Equation 4)

Step 3: Solve the system of equations.

We now have two equations (Equation 3 and Equation 4) with two variables (x and y). We can solve this system of equations using substitution.

From Equation 4, we can express y in terms of z:

y = -7z - 4

Substitute this value of y into Equation 3:

x+6(-7z-4)-10z=30

Simplifying, we get:

x-52z-24-10z=30

x-62z=54

x = 62z + 54

So, the solution to the system of equations is:

x = 62z + 54

y = -7z - 4

z = z (z is a free variable)

This means that for any value of z, we can find the corresponding values of x and y using the given equations.

User Jesselle
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