Solve the system
-3x+4y=-6
5x-3y=-22
3y+2z=-1
I'm not sure how to approach this question because all of the equations are missing a variable??
actually it is good when variables are missing, it makes it easier.
your first two are just in x and y, so solve these two.
first one times 3 ---> -9x + 12y = -18
2nd one times 4 ---> 20x - 12y = -88
add them
11x = -106
x = -106/11
then -3(-106/11) + 4y = -6
y = -96/11
in the third:
3(-96/11) + 2z = -1
z = 277/22
crazy answers, but they work.
Are you sure you typed them correctly?
OMG, sorry reiny, i didn't think anyone was going to help me out on this one so i reposted it (again with the wrong equation)... the correct one is..
-3x=4y=-6
5x-3z=-22 (3z instead of 3y)
3y+2z=-1
And now you made a typo in the first , lol
OK, I think you meant ....
-3x+4y=-6
5x-3z=-22
3y+2z=-1
2nd by 2 ---> 10x - 6z = -44
3rd by 3 ---> 9y + 6z = -3
add them: 10x + 9y = -47 (#4)
1st by 10 ---> -30x + 40y = -60
4th by 3 ----> 30x + 27y = -141
add them: 67y = -201
y = -3 , YEA!!
now backsubstitute.....
Thank you very much Reiny, and Goodnight :) no more questions for now :)
To solve the system of equations, you can use various methods such as substitution, elimination, or matrix methods. Let's solve the given system of equations using the elimination method.
First, let's eliminate one variable from two of the equations. We will eliminate the variable "y" in this case.
From the first equation, -3x + 4y = -6, we can rewrite it as:
4y = 3x - 6 (by adding 3x to both sides)
Similarly, from the second equation, 5x - 3y = -22, we can rewrite it as:
5x = 3y - 22 (by adding 3y to both sides)
Now we have two equations that both have the variable "x," so we can combine them to eliminate the "x" variable. To do this, we need to multiply the first equation by 5 and the second equation by 4:
20y = 15x - 30 (multiplying the first equation by 5)
20x = 12y - 88 (multiplying the second equation by 4)
Now we can subtract the equations to eliminate the "x" variable:
20x - 15x = 12y - 15x - 88 + 30
5x = -3y - 58
At this point, we have an equation relating "x" and "y." Let's label it as Equation 1.
Now let's work on eliminating the "y" variable from Equation 1 and the third equation, 3y + 2z = -1. To do this, we need to multiply the third equation by 2:
6y + 4z = -2
We can then add Equation 1 and the modified third equation to eliminate the "y" variable:
5x + 6y + 4z = -3y - 58 - 2
5x + 10z = -3y - 60 (adding the equations)
Now we have an equation relating "x" and "z." Let's label it as Equation 2.
At this point, we have two equations:
Equation 1: 5x + 10z = -3y - 60
Equation 2: 5x = -3y - 58
Since Equation 2 is already solved for "x," we can substitute it into Equation 1:
-3y - 58 + 10z = -3y - 60
Now we can simplify and solve for "z":
10z - 58 = -60
10z = -60 + 58
10z = -2
z = -2/10
z = -1/5
Now that we have found the value of "z," we can substitute it back into Equation 2 to solve for "x":
5x = -3y - 58
5x = (-3)(-1/5) - 58
5x = 3/5 - 58
5x = 3/5 - 290/5
5x = -287/5
x = -287/25
Finally, substituting the values of "x" and "z" into Equation 1, we can solve for "y":
5x + 10z = -3y - 60
(5)(-287/25) + (10)(-1/5) = -3y - 60
-1435/25 - 2 = -3y - 60
-1435/25 - 50/25 = -3y
-1485/25 = -3y
-297/5 = -3y
y = -(-297/5)/3
y = 297/15
y = 99/5
Therefore, the solution to the given system of equations is x = -287/25, y = 99/5, and z = -1/5.