Linear equation: Solve x1, x2, x3, in terms of y1,y2,y3 x1 + ax2 + bx3 = y1
x2 + bx3 = y2
x3 = y3

x1 = y1 - a*y2 + (a*c - b)*y3

x2 = y2 - c*y3

x3 = y3

Step-by-step explanation:

Here we have the system:

x1 + a*x2 + b*x3 = y1

x2 + c*x3 = y2

x3 = y3

Where the variables y1, y2, and y3 are known (a, b and c are also known).

The first step is to isolate one of the variables in one of the equations, we can see that in the third equation we have x3 already isolated, so now we can just replace it on the other two equations to get:

x1 + a*x2 + b*(y3) = y1

x2 + c*(y3) = y2

Now we again want to isolate one of the variables in one of the equations, i will isolate x2 in the second equation to get:

x2 = y2 - c*y3

Now we can replace this on the other equation to get:

x1 + a*(y2 - c*y3) + b*y3 = y1

Now we canw write x1 in terms of the known variables:

x1 = y1 - a*y2 + (a*c - b)*y3

And in the process we also found that:

x3 = y3

x2 = y2 - c*y3

Then the solutions are:

x1 = y1 - a*y2 + (a*c - b)*y3

x2 = y2 - c*y3

x3 = y3

twenty-six miles over two gallons because the proportion for the original 13 miles to one gallon would be written 13/1 and, doing the same thing to the top and bottom, i multiplied by 2 and got 26/2