Question

I don’t really understand this type of stuff so please help.

Answer 1

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 $y=5x-1$, then when $x=0$, you get $y=-1$; when $x=1$, you get $y=4$. So plot the points (0, -1) and (1, 4), then strike a line through.

1. Notice that dividing both sides of $2y=10x-2$ by 2 returns $y=5x-1$, 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 $x$ or $y$, you can always find a corresponding value for the other variable.)

2. See attached image. $3x-y=2$ 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 $x+3y=9$, so there will be an infinite number of solutions.

4. (by substitution) Since $y=-7x+3$, we can replace $y$ in the second equation:

$-7x+3+7x=10\implies3=10$

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

5. (by substitution) Since $x=2y+2$, in the first equation we have

$-5(2y+2)+3y=-10y-10+3y=-7y-10=11\implies-7y=21\implies y=-3$

Then back in the second equation we find

$x=2(-3)+2=-6+2=-4$

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.