find the point(s) of intersection of the equations?

Question

1. 3x + 4y = 8 and x + y = 5


2. x - y + 1 = 0 and y - x^2 = 7

Answer 1

try and make x or y the subject of the formula
so use x + y = 5 since its easier to make x or y the subject:
i'll make x the subject of the formula so i get
x = 5 - y
now substitute this into the first equation so you get
3(5 - y) + 4y = 8
15 - 3y +4y = 8y = -7

now substitute this into the second equation to get x

x = 5 - y
x = 5 - (-7)
x = 12

so the point of intersection for this line is (12,-7)

2.do the same for the second one

make y the subject of the formula
y = x + 1


now substitute into second equation
x + 1 - x^2 = 7
you can rewrite this into:

x^2 - x+6 = 0

factor this by using quadratic formula:

[-(-1) + ((-1)^2 - (4*1*6)) ^(1/2)] / (2*1)

and

[-(-1) - ((-1)^2 - (4*1*6))^(1/2)] / (2*1)]

since we cannot sqrt a negative there is no solution to the given equation meaning the two graphs do not intersect

Answer 2

3x + 4y = 8
x + y = 5 solve x in term of y so x = 5- y

put x = 5-y for x in the first equation then

3(5-y)+4y = 8
15 -3y +4y = 8
15+y = 8
y= -7 cuz subtract both sides by 15

when we have y = -7 put in back on x = 5 -y
so x = 5 - - 7 = 5+7 =12
solution is x = 12 and y = -7

problem number 2

x - y + 1 = 0
y - x^2 = 7 solve y in term of x so y = 7+x^2

put y = x^2+7 into the first equation then
x - (x^2+7)+1 = 0
x -x^2-7+1 = 0
-x^2+x -6 = 0
time -1 to both sides
x^2-x+6 = 0 no real solutions .
make sure you typed in the right question on number 2.
i think that is y +x^2 = 7 . it is not y -x^2 = 7 , i think you did type wrong.

if you mean y+x^2 = 7 is what you meant then

x - y + 1 = 0
y +x^2 = 7 solve y in term of x so y = 7-x^2

put y = 7 -x^2 into the first equation then
x - (7-x^2)+1 = 0
x -7+x^2+1 = 0
x^2+x -6 = 0
(x+3)(x-2) = 0
x+3= 0 or x-2 = 0
x= -3 and x = 2
put 2 values of x into y = 7-x^2 to figure out y

y = 7-(-3)^2 = 7 -9 = -2

y = 7-(2)^2 = 7-4 = 3

answer is x = -3, y = -2 and x= 2, y= 3

make sure you check your question again to be sure i am solved on the right equation.

Answer 3

Question 1
3x + 4y = 8
- 3x - 3y = - 15-----ADD
y = - 7

x - 7 = 5
x = 12

x = 12 , y = - 7

Question 2
y = x² + 7
x - x² - 7 + 1 = 0
x² - x + 6 = 0
x = [ 1 ± √(1 - 24) ] / 2
x = [1 ± √( - 23) ] / 2
x = [1 ± i √(23) ] / 2