Solve the system of equations algebraically. 1. y=x+2, y=-3x 2. y=x+20, y=6xme

1.

y = x + 2 , y = -3x

Since both are equal to y we can set them equal to each other:

x + 2 = -3x
2 = -4x
-1/2 = x

Now we will substitute the value of x in one of the equation to find the value of the y:

y = x + 2
y = -1/2 + 2
y = 1.5

So the solution of the first system is (-0.5,1.5).

2.

y = x + 20 , y = 6x

Since both are equal to y, we can set them equal to each other:

x + 20 = 6x
20 = 5x
4 = x

Now we will substitute the value of x in one of the equations to find the y:

y = 6x
y = 6(4)
y = 24

So the solution for the second set of equations is (4,24).

Hope this helps :)

Letter b sorry i’m i’m wrong xx

I'll do the first one for you. it isn't very hard at all.on number one, you have (50 × 3), which equals 150, and (3 × 3), which equals 9. after you establish that, you add 150 and 9 together. so, (150 + 9) equals 159.this means that 53 × 3 = 159.