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Math  /  Algebra

Question7. 3w+x+y+z=2x+2y+3z=1−2z=23w2x+4y+7z=1\begin{array}{r}3 w+x+y+z=2 \\ x+2 y+3 z=1 \\ -2 z=2 \\ 3 w \quad 2 x+4 y+7 z=1\end{array}

Studdy Solution

STEP 1

1. We are given a system of four linear equations with four variables: w,x,y, w, x, y, and z z .
2. The goal is to find the values of these variables that satisfy all the equations simultaneously.
3. We will use substitution or elimination methods to solve the system.

STEP 2

1. Simplify and solve one of the equations for one variable.
2. Substitute the solution into the other equations.
3. Solve the resulting system of equations.
4. Back-substitute to find the remaining variables.
5. Verify the solution by substituting all variables back into the original equations.

STEP 3

Let's start with the third equation:
−2z=2 -2z = 2
Solve for z z by dividing both sides by −2-2:
z=−1 z = -1

STEP 4

Substitute z=−1 z = -1 into the other equations:
1. 3w+x+y+(−1)=2 3w + x + y + (-1) = 2 becomes 3w+x+y−1=2 3w + x + y - 1 = 2
2. x+2y+3(−1)=1 x + 2y + 3(-1) = 1 becomes x+2y−3=1 x + 2y - 3 = 1
3. 3w−2(−1)=2 3w - 2(-1) = 2 becomes 3w+2=2 3w + 2 = 2
4. 2x+4y+7(−1)=1 2x + 4y + 7(-1) = 1 becomes 2x+4y−7=1 2x + 4y - 7 = 1

STEP 5

Simplify the equations after substitution:
1. 3w+x+y=3 3w + x + y = 3
2. x+2y=4 x + 2y = 4
3. 3w=0 3w = 0
4. 2x+4y=8 2x + 4y = 8

STEP 6

From equation 3: 3w=0 3w = 0 , solve for w w :
w=0 w = 0

STEP 7

From equation 2: x+2y=4 x + 2y = 4 , solve for x x :
x=4−2y x = 4 - 2y
Substitute x=4−2y x = 4 - 2y into equation 4:
2(4−2y)+4y=8 2(4 - 2y) + 4y = 8

STEP 8

Simplify and solve for y y :
8−4y+4y=8 8 - 4y + 4y = 8
This simplifies to 8=8 8 = 8 , which is always true, indicating that y y can be any real number. Let's choose y=0 y = 0 for simplicity.

STEP 9

Substitute y=0 y = 0 back into the equation for x x :
x=4−2(0)=4 x = 4 - 2(0) = 4

STEP 10

Verify the solution w=0,x=4,y=0,z=−1 w = 0, x = 4, y = 0, z = -1 by substituting back into the original equations:
1. 3(0)+4+0−1=2 3(0) + 4 + 0 - 1 = 2 checks out.
2. 4+2(0)+3(−1)=1 4 + 2(0) + 3(-1) = 1 checks out.
3. 3(0)−2(−1)=2 3(0) - 2(-1) = 2 checks out.
4. 2(4)+4(0)+7(−1)=1 2(4) + 4(0) + 7(-1) = 1 checks out.
All equations are satisfied.
The solution is:
w=0,x=4,y=0,z=−1 w = 0, \quad x = 4, \quad y = 0, \quad z = -1

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