Solve the system using any algebraic method.
2x - 2y + z = 3
5y - z = -31
x + 3y + 2z = -21
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The answer I got for this is ( -26, 6,67) but I know its wrong and I don't know why. Help please >< my work is at the bottom
I used substitution method changed 5y-z=-31 to z=5y+31 then I:
Plugged it in the first equation=
2x-2y+5y+31=3 -> 2x+3y+31=3 -> 2x+3y=-28 is my new 1st equation...
Plugged in z=5y+31 to equation 3 =
x+3y+2(5y+31)=-21 -> x+3y+10y+62=-21 -> x+13y+62=-21 -> x+13y=-83 is my new 2nd equation.
2(=13y=82)+3y=-28 -> -26-166+3y=-28 -> -23y=138 - > y=6
2x+3(6)=-28 -> 2x+18=-28 -> 2x=-46 -> x=-23
2(-26)-2*6)+z=3 -> -52-12+z=3 -> -64+z=3 -> z=67
5y - z = -31
z = 5y + 31
2x - 2y + z = 3
( Now put z = 5y + 31 in equation )
2x - 2y + 5y + 31 = 3
2x + 3y = 3 - 31
2x + 3y = - 28
2x = - 28 - 3y Divide both sides with 2
x = - 14 - ( 3 / 2 ) y
x + 3y + 2z = -21
( Now put x = - 14 - ( 3 / 2 ) y and z = 5y + 31 in equation )
- 14 - ( 3 / 2 ) y + 3y + 2 ( 5y + 31 ) = - 21
- 14 - ( 3 / 2 ) y + 3y + 10y + 62 = - 21
- ( 3 / 2 ) y + 13y = - 21 + 14 - 62
- ( 3 / 2 ) y + ( 26 / 2 ) y = - 69
( 23 / 2 ) y = - 69 Multiply both sides with 2
23y = - 138 Divide both sides with 23
y = - 6
z = 5y + 31
z = 5 ( -6 ) + 31
z = - 30 + 31
z = 1
x = - 14 - ( 3 / 2 ) y
x = - 14 - ( 3 / 2 ) -6
x = - 14 + 18 / 2
x = - 14 + 9
x = - 5
So solutions are :
x = - 5
y = - 6
z = 1
Your substitutions were correct, but there seems to be a mistake in your calculations. Let's go through the steps again to find the correct solution.
The original system of equations is:
1) 2x - 2y + z = 3
2) 5y - z = -31
3) x + 3y + 2z = -21
Start by solving equation 2) for z:
5y - z = -31
z = 5y + 31
Now substitute this value of z into equations 1) and 3):
1) 2x - 2y + (5y + 31) = 3
2) x + 3y + 2(5y + 31) = -21
Simplify these equations:
1) 2x + 3y = -28
2) x + 13y = -83
Now you have a system of two equations with two variables. You can solve this system using any method, such as substitution or elimination.
Let's use substitution method to solve this system. Solve equation 1) for x:
2x = -28 - 3y
x = (-28 - 3y)/2
Now substitute this value of x into equation 2):
((-28 - 3y)/2) + 13y = -83
Multiply both sides by 2 to eliminate the fraction:
-28 - 3y + 26y = -166
Combine like terms:
23y = -138
Divide both sides by 23:
y = -6
Now substitute this value of y back into equation 1) to solve for x:
2x + 3(-6) = -28
2x - 18 = -28
2x = -10
x = -5
Finally, substitute the values of x and y into equation 2) to solve for z:
z = 5y + 31 = 5(-6) + 31 = -30 + 31 = 1
Therefore, the correct solution to the system of equations is x = -5, y = -6, z = 1.
To solve this system of equations, you can use the method of substitution or elimination. Let's use the method of substitution.
Start with the first equation:
2x - 2y + z = 3
Next, solve the second equation for z:
5y - z = -31
Rearrange the equation to get:
z = 5y + 31
Substitute the value of z in the first equation:
2x - 2y + (5y + 31) = 3
Combine like terms:
2x + 3y + 31 = 3
Rearrange the equation to isolate x:
2x + 3y = -28
Now, let's work on the third equation:
x + 3y + 2z = -21
Substitute the value of z from the second equation:
x + 3y + 2(5y + 31) = -21
Simplify:
x + 3y + 10y + 62 = -21
Combine like terms:
x + 13y + 62 = -21
Rearrange the equation to isolate x:
x + 13y = -83
So, the system of equations becomes:
2x + 3y = -28 (Equation 1)
x + 13y = -83 (Equation 2)
Now, you can solve this system of equations using any algebraic method. Let's use substitution.
From Equation 1, isolate x:
2x = -28 - 3y
x = (-28 - 3y)/2
Substitute this value of x into Equation 2:
(-28 - 3y)/2 + 13y = -83
Multiply both sides by 2 to get rid of the fraction:
-28 - 3y + 26y = -166
Combine like terms:
23y = -138
Divide both sides by 23:
y = -6
Substitute the value of y into Equation 1 to solve for x:
2x + 3(-6) = -28
2x - 18 = -28
Add 18 to both sides:
2x = -10
Divide both sides by 2:
x = -5
Finally, substitute the values of x and y into Equation 2 to solve for z:
-5 + 13(-6) = -83
-5 - 78 = -83
-83 = -83
So, the solution to the system of equations is x = -5, y = -6, and z = -83.
It seems like you made a mistake when substituting the value of z into the first equation. The correct substitution would be:
2x - 2y + (5y + 31) = 3
2x - 2y + 5y + 31 = 3
2x + 3y + 31 = 3
2x + 3y = -28
From there, you continued correctly and obtained the correct values for x and y. However, when substituting those values into the equation to find z, it seems like you made a calculation error. The correct calculation should be:
2(-26) - 2(6) + z = 3
-52 - 12 + z = 3
-64 + z = 3
So, the correct value for z is z = 67. Therefore, the correct solution to the system of equations is x = -26, y = 6, and z = 67.