To solve this system, we can substitute the value of z from the second equation into the first equation.
Substituting z = 3w - 2 into w + (1/7)z = 4, we get:
w + (1/7)(3w - 2) = 4
w + (3/7)w - (2/7) = 4
Multiplying both sides of the equation by 7 to eliminate the fractions, we get:
7w + 3w - 2 = 28
10w - 2 = 28
Adding 2 to both sides of the equation, we get:
10w = 30
Dividing both sides of the equation by 10, we get:
w = 3
Now, substitute the value of w back into the second equation to find z:
z = 3w - 2
z = 3(3) - 2
z = 9 - 2
z = 7
Therefore, the solution to the system of equations is w = 3 and z = 7.
Solve this system:
w + 1/7z = 4
z = 3w - 2
7 answers
Is it possible to do it without eliminating the fractions?
Yes, it is possible to solve the system without eliminating the fractions.
Starting from the equations:
w + (1/7)z = 4 ...(1)
z = 3w - 2 ...(2)
We can substitute the expression for z from equation (2) into equation (1):
w + (1/7)(3w - 2) = 4
w + (3/7)w - (2/7) = 4
Multiplying through by 7 to get rid of the fractions gives:
7w + 3w - 2 = 28
10w - 2 = 28
Adding 2 to both sides of the equation,
10w = 30
Dividing both sides by 10 gives:
w = 3
Substituting the found value of w into equation (2):
z = 3w - 2
z = 3(3) - 2
z = 9 - 2
z = 7
Therefore, the solution to the system of equations is w = 3 and z = 7.
Starting from the equations:
w + (1/7)z = 4 ...(1)
z = 3w - 2 ...(2)
We can substitute the expression for z from equation (2) into equation (1):
w + (1/7)(3w - 2) = 4
w + (3/7)w - (2/7) = 4
Multiplying through by 7 to get rid of the fractions gives:
7w + 3w - 2 = 28
10w - 2 = 28
Adding 2 to both sides of the equation,
10w = 30
Dividing both sides by 10 gives:
w = 3
Substituting the found value of w into equation (2):
z = 3w - 2
z = 3(3) - 2
z = 9 - 2
z = 7
Therefore, the solution to the system of equations is w = 3 and z = 7.
Can you solve this system?
3x + 2y - z + 5w = 20
y = 2z - 3w
z = w + 1
2w = 8
3x + 2y - z + 5w = 20
y = 2z - 3w
z = w + 1
2w = 8
To solve this system, we can use a combination of substitution and elimination.
1) From the fourth equation, we can solve for w:
2w = 8
Divide both sides by 2:
w = 4
2) Now substitute w = 4 into the third equation to find z:
z = w + 1
z = 4 + 1
z = 5
3) Next, substitute w = 4 and z = 5 into the second equation to find y:
y = 2z - 3w
y = 2(5) - 3(4)
y = 10 - 12
y = -2
4) Finally, substitute w = 4, z = 5, and y = -2 into the first equation to find x:
3x + 2y - z + 5w = 20
3x + 2(-2) - 5 + 5(4) = 20
3x - 4 - 5 + 20 = 20
3x + 11 = 20
Subtract 11 from both sides:
3x = 9
Divide both sides by 3:
x = 3
Therefore, the solution to the system of equations is:
x = 3
y = -2
z = 5
w = 4
1) From the fourth equation, we can solve for w:
2w = 8
Divide both sides by 2:
w = 4
2) Now substitute w = 4 into the third equation to find z:
z = w + 1
z = 4 + 1
z = 5
3) Next, substitute w = 4 and z = 5 into the second equation to find y:
y = 2z - 3w
y = 2(5) - 3(4)
y = 10 - 12
y = -2
4) Finally, substitute w = 4, z = 5, and y = -2 into the first equation to find x:
3x + 2y - z + 5w = 20
3x + 2(-2) - 5 + 5(4) = 20
3x - 4 - 5 + 20 = 20
3x + 11 = 20
Subtract 11 from both sides:
3x = 9
Divide both sides by 3:
x = 3
Therefore, the solution to the system of equations is:
x = 3
y = -2
z = 5
w = 4
Thank you!
You're welcome!
1 answer