solve by substitution?

Question

x -3y =15
x^2+y^2=25

Answer 1

x = 15 + 3y

x² = (15 + y).(15 + y)

x² = 225 + 90 y + 9 y²

225 + 90 y + 10y² = 25

10y² + 90y + 200 = 0

y² + 9y + 20 = 0

(y + 5).(y + 4) = 0

y = - 5 , - 4

x = 0 , 3

(0, - 5) , (3, -4)

Answer 2

The solutions are: (i) x = 3, y = - 4, or (ii) x = 0, y = - 5.

Here's how to find them. From the first equation, x = 15 + 3 y, so substitute that into the second equation:

(15 + 3 y)^2 + y^2 = 25. So:

225 + 90 y + 10 y^2 = 25, that is 10 y^2 +90 y +200 = 0, or

10 (y^2 + 9 y +20) = 0 so that (y + 4)(y + 5) = 0.

The solutions are y = - 4 and x = 15 + 3(- 4) = 15 - 12 = 3,

or y = - 5 and x = 15 + 3(- 5) = 0.

CHECK? Yes, these satisfy the original equations!

Live long and prosper.

P.S. Note that the next response is confused/confusing; you don't "back substitute to determine which of the roots is the correct one to get x." It's NOT a question of only one of the roots being "the correct one to get x." There is a value of x for EACH of the values of y, so you simply back substitute to determine each corresponding x.

Answer 3

x-3y=15......(1)
x^2+y^2+25...(2)
From eqn 1 ,we get
x=15+3y
Substituting the value of x in eqn 2 we get
(15+3y)^2+y^2=25
=>225+9y^2+90y+y^2=25
=>10y^2+90y+200=0
=>y^2+9y+20=0
=>(y+4)(y+5)=0
Hence y=-4 or- 5
when y=-4,x=15-12=3
when y= -5,x=15-15=0 Hence x= 3,y= -4 and x=0,y= -5 ans

Answer 4

Substitute 3y + 15 for x in the 2,nd equation to get
(3y + 15)² + y² = 15 multiply it out and simplify to get a quadratic in y. Solve the quadratic and back substitute to determine which of the roots is the correct one to get x.

HTH ☺


Doug

Answer 5

start with
x-3y=15
-3y=15-x (subtract x from both sides)
y=(15-x)/-3 (divide by -3, both sides)
so, y=-5+x/3...put this into the second equation

x^2+(-5+x/3)^2=25...solve for x
x^2+(-5+x/3)*(-5+x/3)=25
x^2+25-5x/3-5x/3+x^2/9=25 simplify
x^2-10x/3+x^2/9=0 multiply by 9
9x^2-30x+x^2=0
10x^2-30x=0
x*(10x-30)=0
x=0 and
10x-30=0
x=3
so, x=0 or x=3,
plug these x values in to either equation to find y

i choose the easy one, the first equation

when x=0
0-3y=15
-3y=15
y=-5

when x=3
3-3y=15
-3y=12
y=-4

solved when (x=0 and y=-5) or (x=3 and y=-4)....

Hope this is clear?

Answer 6

ok x=15+3y
so substitute
(15+3y)^2+y^2=25
225+90y+9y^2=25
9y^2+90y+200=0
and go from there to tired to do the rest sorry