Question 27

If $x, y, z$ satisfy the system of equations

3x + 2y - z = 3

2x + y - 3z = 4

x - y + 2z = -1

Then the value of $2x + 2y - 3z =$ ....

Given the system of equations

3x + 2y - z = 3 \quad ...(1)

2x + y - 3z = 4 \quad ...(2)

x - y + 2z = -1 \quad ...(3)

Multiply the first equation by $3$ and the second equation by $1$ , then subtract.

3x + 2y - z = 3 \quad |\times 3| \quad 9x + 6y - 3z = 9

2x + y - 3z = 4 \quad |\times 1| \quad 2x + y - 3z = 4 \quad -

7x + 5y = 5 \quad ...(4)

Multiply the first equation by $2$ and the third equation by $1$ , then subtract.

3x + 2y - z = 3 \quad |\times 2| \quad 6x + 4y - 2z = 6

x - y + 2z = -1 \quad |\times 1| \quad x - y + 2z = -1 \quad -

7x + 3y = 5 \quad ...(5)

Subtract the fifth equation from the fourth equation to find the value of $y$ .

7x + 5y = 5

7x + 3y = 5 \quad -

2y = 0

y = 0

Substitute $y = 0$ into the fourth equation to find the value of $x$ .

7x + 5y = 5

7x + 5(0) = 5

7x = 5

x = \frac{5}{7}

Substitute $x = \frac{5}{7}$ and $y = 0$ into the third equation to find the value of $z$ .

x - y + 2z = -1

\frac{5}{7} - 0 + 2z = -1

2z = -\frac{12}{7}

z = -\frac{6}{7}

Substitute the values $x = \frac{5}{7}$ , $y = 0$ , and $z = -\frac{6}{7}$ into $2x + 2y - 3z$ .

2x + 2y - 3z = 2\left(\frac{5}{7}\right) + 2(0) - 3\left(-\frac{6}{7}\right)

2x + 2y - 3z = \frac{10}{7} - 0 + \frac{18}{7}

2x + 2y - 3z = \frac{28}{7}

2x + 2y - 3z = 4

Therefore, the value of $2x + 2y - 3z$ is $4$ .

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