Question

Can someone teach me how to solve this problem please:)

Answer 1

y=-5, y=3

x=-3., x=5

Explanation:

x^2+y^2=34

3x-3y=6

isolate x in te equation

3x-3y=6

x=3/3 y+6/3

x=y+2

plug the y+2 in the equation:

x^2+y^2=34

(y+2)^2+y^2=34

y^2+4y+4+y^2=34

2y^2+4y=34-4

2y^2+4y=30 divide by 2

y^2+2x-15=0 factorize

(y+5)(y-3)=0 eiter y+5=0 ten y=-5 or y-3=0 then y=3

now plug the solution in the equation

3x-3y=6

3x-3(-5)=6

3x=6-15

x=-9/3=-3

for y=3

3x-3y=6

3x-9=6

3x=15

x=5

Answer 2

x= -3, y= -5

or x= 5, y=3

Explanation:

① Label the 2 equations

x² +y²= 34 -----(1)

3x -3y= 6 -----(2)

From (2):

x -y= 2 -----(3)

Notice that (x-y)²= x² -2xy +y²

Thus, (equation 3)²= (equation 1) -2xy

Squaring (3):

(x -y)²= 2²

(x -y)²= 4

Expand terms in bracket:

x² -2xy +y²= 4

x² +y² -2xy= 4 -----(4)

subst. (1) into (4):

34 -2xy= 4

2xy= 34 -4 (bring constant to 1 side)

2xy= 30 (simplify)

xy= 30 ÷2 (÷2 throughout)

xy= 15 -----(5)

From (3):

x= y +2 -----(6)

I'll rewrite 2 of the equations.

x= y +2 -----(6)

xy= 15 -----(5)

Subst. (6) into (5):

y(y+2)= 15

y² +2y= 15

y² +2y -15= 0

(y +5)(y -3)=0

y+5= 0 or y-3=0

y= -5 or y= 3

Subst. into (6):

x= -5 +2 or x= 3 +2

x= -3 or x= 5