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

answer: c, the green line

step-by-step explanation:

there is no black line, eliminate b.

simple interest implies periodic payments, but the red line shows no payments, eliminate d.

simple interest has equal payments. if paid say every day, the graph is about a straight line (actually a stairstep).

the yellow line is not straight, eliminate a.

check: the green line is straight, answer is c.

simple interest with positive interest rate gives a straight line sloping up to the right. green line:

g(t) = $100 + ($200-$100) ÷ (20 years) × (t years)

= $100 + (4%/year) × ($100)(t years)

incidentally, the yellow line curving upward represents continuously compounded interest

y = $100 × 2^(t years / (14 assuming the yellow line hits $200 at 14 years. 2^0 is 1, so y(0) = $100; 2^(14/14) is 2, so y(14) = $200.