1) 2a+c=-10
b-c=15
a-2b+c=-5
2) x+y+z=3
13x+2x=2
-x-5z=-5
3)x-y+3z=3
-2x+2y-6z=6
y-5z=-3
4)3x-2y+z=1
-x+y-z=2
5x+2y+10z=-39
5)3x-2y+z=1
-x-y-z=-3
3x-2y+3z=-12
http://www.wolframalpha.com/input/?i=2a%2Bc%3D-10%2C+b-c%3D15%2C+a-2b%2Bc%3D-5+
2a+c=-10, b-c=15, a-2b+c=-5
To solve each system of equations, we can use the method of elimination or substitution. Here, I'll explain how to solve the first system of equations to help get you started.
1) 2a + c = -10
b - c = 15
a - 2b + c = -5
Let's use the method of substitution:
Step 1: Take the second equation, b - c = 15, and solve it for b.
b = 15 + c
Step 2: Substitute the value of b from Step 1 into the other equations:
2a + c = -10 (1)
a - 2(15 + c) + c = -5 (3)
Simplify equation (3):
a - 30 - 2c + c = -5
a - c = 25 (4)
Step 3: Now, we have two equations (1) and (4):
2a + c = -10 (1)
a - c = 25 (4)
From equation (4), solve for a:
a = 25 + c (5)
Substitute this value of a from equation (5) into equation (1):
2(25+c) + c = -10
50 + 2c + c = -10
3c = -60
c = -20
Step 4: Substitute the found value of c into equation (5) to get the value of a:
a = 25 + (-20)
a = 5
Step 5: Substitute the found values of a and c into one of the original equations to solve for b:
b - c = 15
b - (-20) = 15
b + 20 = 15
b = 15 - 20
b = -5
Therefore, the solution to the first system of equations is:
a = 5, b = -5, c = -20.
You can follow these steps to solve the rest of the systems of equations as well.