solve systems of equations using the substitution method?

1) Since they both are expressed as y we can say that:

x^2+4x+3=2x+6 subtract 2x from both sides

x^2+2x-3=0 factor...

(x+3)(x-1)=0, so x=1 and -3 and since y=2x+6, y=8 and 0

So the two solutions to the system of equations are the points (1,8) and (-3,0)

...

2.) y=x^2-6x+8 and 2y+x=4, substituting the y from equation 1 into equation 2 gives

2(x^2-6x+8)+x=4

2x^2-12x+16+x=4

2x^2-11x+12=0... factor...

2x^2-8x-3x+12=0

2x(x-4)-3(x-4)=0

(2x-3)(x-4)=0, so x=1.5 and 4, and since 2y+x=4 we find

y=(4-x)/2, y=1.25 and 0 so the solutions to this system of equations are the points

(1.5, 1.25) and (4, 0)

...

3.) .y=-x^2+2x+4 and x+y=4, so y=4-x, now substitute this into the first equation to get:

4-x=-x^2+2x+4

x^2-3x=0 factor

x(x-3)=0, so x=0 and 3, and since y=4-x:

y=4 and 1, so the solutions to this system of equations are the points (0, 4) and (3, 1)

...

4.) y=x^2+2x+7 and y=x+7 so we can just use the value of y from the 2nd in the 1st...

x+7=x^2+2x+7

x^2+x=0 factor

x(x+1)=0, so x=-1 and 0, and since y=x+7:

y=6 and 7, so the solutions to the system of equations are the points:

(-1, 6) and (0, 7)

Too many questions !

Will do number 1

x² + 4x + 3 = 2x + 6

x² + 2x - 3 = 0

(x + 3)(x - 1) = 0

x = - 3 , x = 1

y = 0 , y = 8

these are all quadratics and you certainly can solve them yourself............

#1 [ 2x + 6 = x² + 4x + 3 ] ---> x² + 2x - 3 = 0...........[ x+3][x-1] = 0

[