10x+10y+20=1850
20x+20y+10z=2200
20x+10y+10z=1500
I bet you mean
10x+10y+20z=1850
20x+20y+10z=2200
20x+10y+10z=1500
get rid of x and y in first two by multiplying first eqn by 2
20x+20y+40z=3700
................> 30 z = 1500 so z = 50
20x+20y+10z=2200
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then third and first
20x+10y+10z=1500 >> 20 x + 10 y = 1000
20x+20y+40z=3700 >> 20 x + 20 y = 1700
so
-20 y = -700
y = 35
go back and get x from anywhaere knowing y is 35 and z is 50
To solve this system of equations, we can use the method of substitution. This method involves solving one equation for one variable and substituting it into the other equations. Let's begin.
Equation 1: 10x + 10y + 20 = 1850
Equation 2: 20x + 20y + 10z = 2200
Equation 3: 20x + 10y + 10z = 1500
Let's solve Equation 1 for one variable. We can isolate either x or y. Here, let's solve for x:
10x + 10y + 20 = 1850
10x = 1850 - 10y - 20
10x = 1830 - 10y
x = (1830 - 10y)/10
x = 183 - y
Now, substitute this expression for x into the other equations to solve for y and z.
Substitute x in Equation 2:
20(183 - y) + 20y + 10z = 2200
3660 - 20y + 20y + 10z = 2200
10z = 2200 - 3660
10z = -1460
z = -146
Substitute x in Equation 3:
20(183 - y) + 10y + 10z = 1500
3660 - 20y + 10y + 10z = 1500
-10y + 10z = 1500 - 3660
-10y + 10z = -2160
-10y = -2160 - 10z
y = (2160 + 10z)/10
y = 216 + z
Now we've found expressions for x, y, and z in terms of z. We can choose any value for z and find the respective values for x, y, and z.
For example, let's choose z = 0:
x = 183 - y = 183 - (216 + 0) = 183 - 216 = -33
y = 216 + z = 216 + 0 = 216
z = 0
The solution to the system of equations is x = -33, y = 216, and z = 0.