Idon’t really understand this type of stuff so .

For 1 and 2, you plot both lines, and wherever they intersect is the solution to the system. Given the equation of a line, I think the easiest way to plot it is to find two points on the line, then draw a line through them. For example, if , then when , you get ; when , you get . So plot the points (0, -1) and (1, 4), then strike a line through.

1. Notice that dividing both sides of by 2 returns , same as the first equation. So the system of equations reduces to one equation, which can have an infinite number of solutions. (This is because for any choice of or , you can always find a corresponding value for the other variable.)

2. See attached image. is given by the purple line.

For 3-6, you have several options. The two simplest methods of solving them are by substitution or elimination.

3. Like with (1), notice that dividing both sides of the first equation by 2 gives , so there will be an infinite number of solutions.

4. (by substitution) Since , we can replace in the second equation:

but this is false, so there are no solutions to this system.

5. (by substitution) Since , in the first equation we have

Then back in the second equation we find

So (-4, -3) is the only solution here.

6. (by substitution) Notice that the left hand sides of both equations are the same, so we end up with 7 = 12, but this is false, so no solution exists.

this is the formula: v = the value of investment at the end of the time period

p = the principal amount (the initial amount invested)

r = the annual interest rate

n = the annual frequency of compounding (how many times a year interest is added)

t = the number of years the money is invested

^ means raise to the power of