Can u help me with this simultaneous equation?

Question

X2 + Y2 = 29
Y-X = 3

where X2 is x squared and Y2 is y squared

Answer 1

y-x=3 (eqn1)
x^2+y^2=29 (eqn 2)

re arrange eqn 1
x=y-3

no substitute into eqn 2
(y-3)^2+y^2=29

Expand the brackets:
y^2-6y+9+y^2=29

Collect Terms:
2y^2-6y-20=0

y^2-3y-10=0

Factorise
(y-5)(y+2)=0

therefore y=5 or y=-2

sub those values into eqn 1:
x=5-3=2
or x=-2-3=5

so: y=5 or y=-2
and: x=2, or x=-5

if you want to double check your answer put your values into either eqn and the right answers should come out.

Answer 2

Now I can see why algebra was always a mystical subject to me - I can't even do a simple equation.

Tell your maths teacher to get his head around this!

If X squared + Y squared = 29 then the two numbers added together must = 29

So if Y - X = 3 then Y is 2 otherwise the answer would be - something, which means that Ysquared = 4 and that means that
X squared is the difference.

Now I'm retired I realise that algebra was surplus to requirements and it would have been more useful for me to have learnt about balance sheets!!!

Good Luck

Answer 3

Sure, the first equation defines a circle, and the second equation is a straight line. The circle has a radius over 5 and the line has a y-intercept of 3, so I am fairly sure the line will intersect the circle at two points. Substitute y=x+3 into the equation of the circle, find the 2 x co-ordinates and then substitute back to find the y co-ordinates.

Answer 4

solve for x or y in the second equation and plug it in to the first equation, and solve... i'll solve for y
y-x=3 (add x to both sides)
y = x +3
so,
x^2 + (x + 3)^2 = 29 - remember (x+3)^2 = (x+3)(x+3)
x^2 + x^2 + 3x + 3x + 9 = 29
simplify it, and...
2x^2 + 6x +9 = 29
subtract 9 from both sides...
2x^2 + 6x = 20
divide the whole thing by two (aka factor it out)
x^2 + 3x = 10
make it equal to zero
x^2 + 3x -10 = 0
factor
(x+5)(x-2) = 0
set each factor equal to zero, and x = -5; 2

Answer 5

solve for y = 3+x

Subs into (1)

x2 + (3+x)2 = 29

expand

solve for x using the quadratic formula.

QED

Answer 6

x^2+y^2 =29..........(1)
x-y=3....................(2)
from (2),x=y+3
now substitute into (1)
(y+3)^2+y^2 = 29
y^2+6y+9+y^2-29 =0
2y^2+6y-20=0
y^2+3y-10=0
factorise
(y-2)(y+5)=0
>>>>y=2 or -5

substitute into (2)
when y=2,x=5
when y= -5,x= -2

i hope that this helps

Answer 7

Nope, sorry