#1. y=x^2+4x+3
y= 2x+6
#2.y=x^2-6x+8
2y+x=4
#3.y=-x^2+2x+4
x+y=4
#4. y=x^2+2x+7
y=x+7
for each i need to know
1) find the value(s) of x
2) find the value(s) of y
3) write the solution(s) in coordinate form
Thomas
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)
Como
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
ted s
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
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