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Math  /  Algebra

QuestionSolve the given system of equations 5x+2y−5z=−62x−4y+2z=−44x−5y+4z=−17\begin{array}{l} 5 x+2 y-5 z=-6 \\ 2 x-4 y+2 z=-4 \\ 4 x-5 y+4 z=-17 \end{array}

Studdy Solution

STEP 1

1. We are given a system of three linear equations with three variables: xx, yy, and zz.
2. The goal is to find the values of xx, yy, and zz that satisfy all three equations simultaneously.
3. We will use the method of elimination to solve this system.

STEP 2

1. Eliminate one variable from two pairs of equations.
2. Solve the resulting two-variable system.
3. Substitute back to find the third variable.
4. Verify the solution by substituting back into the original equations.

STEP 3

Choose two pairs of equations to eliminate the same variable. Let's eliminate zz.
First, consider the first and second equations: \begin{align*} 5x + 2y - 5z &= -6 \\ 2x - 4y + 2z &= -4 \end{align*}
Multiply the second equation by 2.52.5 to align the coefficients of zz: 2.5(2x−4y+2z)=2.5(−4)2.5(2x - 4y + 2z) = 2.5(-4) 5x−10y+5z=−105x - 10y + 5z = -10
Now add this to the first equation: \begin{align*} 5x + 2y - 5z &= -6 \\ 5x - 10y + 5z &= -10 \end{align*}
Add the equations: 10x−8y=−1610x - 8y = -16

STEP 4

Now, eliminate zz from the first and third equations: \begin{align*} 5x + 2y - 5z &= -6 \\ 4x - 5y + 4z &= -17 \end{align*}
Multiply the first equation by 44 and the third by 55: 4(5x+2y−5z)=4(−6)4(5x + 2y - 5z) = 4(-6) 5(4x−5y+4z)=5(−17)5(4x - 5y + 4z) = 5(-17)
This gives: 20x+8y−20z=−2420x + 8y - 20z = -24 20x−25y+20z=−8520x - 25y + 20z = -85
Add the equations: 45y=−10945y = -109

STEP 5

Solve for yy: y=−10945y = \frac{-109}{45}
Substitute yy back into the equation from STEP_1: 10x−8(−10945)=−1610x - 8\left(\frac{-109}{45}\right) = -16
Simplify and solve for xx: 10x+87245=−1610x + \frac{872}{45} = -16
Multiply through by 45 to clear the fraction: 450x+872=−720450x + 872 = -720
Solve for xx: 450x=−720−872450x = -720 - 872 450x=−1592450x = -1592 x=−1592450x = \frac{-1592}{450}

STEP 6

Simplify xx: x=−796225x = \frac{-796}{225}

STEP 7

Substitute xx and yy back into one of the original equations to solve for zz. Use the first equation: 5(−796225)+2(−10945)−5z=−65\left(\frac{-796}{225}\right) + 2\left(\frac{-109}{45}\right) - 5z = -6
Simplify and solve for zz: −3980225+−21845−5z=−6\frac{-3980}{225} + \frac{-218}{45} - 5z = -6
Multiply through by 225 to clear fractions: −3980−1090−1125z=−1350-3980 - 1090 - 1125z = -1350
Solve for zz: −1125z=2720-1125z = 2720 z=27201125z = \frac{2720}{1125}

STEP 8

Simplify zz: z=544225z = \frac{544}{225}

STEP 9

Verify the solution by substituting xx, yy, and zz back into the original equations.
Check each equation to ensure both sides are equal.
The solution is: x=−796225,y=−10945,z=544225x = \frac{-796}{225}, \quad y = \frac{-109}{45}, \quad z = \frac{544}{225}

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