Question

Solve by elimination
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Answer 1

`\begin{cases}x = 1 & \text{} \\y = 1 & \text{} \\ z = 0& \text{}\end{cases}`

Explanation:

`\begin{cases}- 2x + 2y + 3z = 0 & \text{} \\- 2x - y + z = - 3 & \text{} \\ \: \: \: 2x + 3y +3z = 5& \text{}\end{cases}`

• Rearrange

`\begin{cases} - 2x + 2y + 3z = 0 & \text{} \\ \: \: \: \: z = - 3 + 2x + y & \text{} \\ \: \: \:2x + 3y + 3z = 5& \text{}\end{cases}`

• Substitute into one of the equations

`\begin{cases} - 2x + 2y + 3( - 3 + 2x + y) = 0 & \text{}\\ \: \: \: 2x + 3y + 3( - 3 + 2x + y)=5& \text{}\end{cases}`

• Apply the Distributive Property

`\begin{cases} - 2x + 2y - 9 + 6x + 3y = 0 & \text{}\\ \: \: \: 2x + 3y - 9 + 6x + 3y= 5&\text{}\end{cases}`

• Combine like terms

`\begin{cases} 4x + 5y - 9 = 0 & \text{}\\ 8x + 6y - 9 = 5& \text{}\end{cases}`

• Rearrange variables to the left side of the equation

`\begin{cases} 4x + 5y = 9 & \text{}\\ 8x + 6y= 14& \text{}\end{cases}`

• Rearrange like terms to the same side of the equation

`\begin{cases}4x= 9 - 5y& \text{}\\ 8x + 6y= 14& \text{}\end{cases}`

• Divide both side of the equations by the coefficient of variable

`\begin{cases} x= (9 - 5y)/(4)& \text{}\\ 8x + 6y= 14& \text{}\end{cases}`

• Substitute into one of the equations

`\bf\longrightarrow{8 * (9 - 5y)/(4) + 6y = 14}`

• Multiply both sides of the equation by the common denominator

`\bf\longrightarrow{8 * (9 - 5y * 4)/(4) + 6y * 4= 14 * 4}`

• Reduce the fractions

`\bf\longrightarrow{8(9 - 5y) + 6y * 4 = 14 * 4}`

• Apply the Distributive Property

`\bf\longrightarrow{72 - 40y + 6y * 4 = 14 * 4}`

• Multiply the monomials

`\bf\longrightarrow{72 - 40y + 24y = 14 * 4}`

• Calculate the product or quotient

`\bf\longrightarrow{72 - 40y + 24y = 56}`

• Combine like terms

`\bf\longrightarrow{72 - 16y = 56}`

• Rearrange variables to the left side of the equation

`\bf\longrightarrow{ - 16y = 56 - 72}`

• Calculate the sum or difference

`\bf\longrightarrow{ - 16y = - 16}`

• Divide both sides of the equation by the coefficient of variable

`\bf\longrightarrow{y = ( - 16)/( - 16) }`

• Calculate

`\bf\longrightarrow{y = 1}`

• Substitute into one of the equation

`\bf\longrightarrow{x = (9 - 5)/(4) }`

• Calculate the sum or difference

`\bf\longrightarrow{x = (4)/(4)}/tex]</h3><ul><li><strong>Cross out the common factor</strong></li></ul><p></p><h3>[tex]\bf\longrightarrow{x = 1}`

• The solution of the system is

`\begin{cases}x = 1 & \text{} \\y = 1 & \text{}\end{cases}`

• Substitute

`\bf\longrightarrow{z = - 3 + 2 + 1}`

• Solve the equation

`\bf\longrightarrow{z = 0}`

• Write the solution

`\begin{cases}x = 1 & \text{} \\y = 1 & \text{} \\ z = 0& \text{}\end{cases}`

Answer 2

(x, y, z) = (1, 1, 0)

Explanation:

You want to solve the given system of equations by elimination.

Solution

There are a couple of nice choices for variables to eliminate.

The last equation has an x-coefficient that is the opposite of the x-coefficient in the other two equations. Adding that equation to those will eliminate the x-variable:

(-2x +2y +3z) +(2x +3y +3z) = (0) +(5) ⇒ 5y +6z = 5 . . . . add eqns 1, 3

(-2x -y +z) +(2x +3y +3z) = (-3) +(5) ⇒ 2y +4z = 2 . . . . add eqns 2, 3

The second of these equations can be divided by 2 to give ...

y +2z = 1

Three times this equation can be subtracted from the first of the reduced equations to eliminate z:

(5y +6z) -3(y +2z) = (5) -3(1) ⇒ 2y = 2

y = 1 . . . . divide by 2

Substitution

Now, we can substitute this value into the previous equation to find the other variables.

y +2z = 1 = 1 +2z ⇒ 0 = 2z ⇒ z = 0

-2x -y +z = -3 = -2x -(1) +(0) ⇒ -2 = -2x ⇒ x = 1

The solution is (x, y, z) = (1, 1, 0).

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Additional comment

The attachment shows a calculator solution to the system of equations by reducing the augmented matrix to "reduced row-echelon form." In general, that is a process of elimination. Each equation ends up with one variable having a coefficient of 1, so we have x = 1, y = 1, z = 0.

You could also start by eliminating the y-variables. Adding twice the second equation to the first will do that, as will adding 3 times the second equation to the third. The result of these operations is two equations in x and z.