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For simultaneous equations, the method is always to write x in terms of y in one equation, which we can name it x(y), then replace the x in another equation with x(y), then solve for y.

Since you are already given the solutions, I would like to focus on the geometric interpretation of the simultaneous equations that you were given. If you find two values for x, these will be the x coordinates of the points of intersection of the straight line and the circle. When you substitute in order to find the values of y, you actually find the y coordinates of the points of intersection.

Some food for thought: What happens if we find only one value for x (thus one value for y as well)?

Think about tangent lines to the circle..... ;-)

x^2 + y^2 = 25    
y - 3x = 13              

Rearrange : add 3x to both sides to give
y = 13 + 3x    

Sub Eq  into :
x^2 + (13 + 3x)^2 = 25
x^2 + 169 + 26x + 9x^2 = 25
10x^2 + 26x + 144 = 0
5x^2 + 13x +72 = 0  by dividing by 2
(5x + 24)(x + 3) = 0  by factorising

Hence two solutions for x are -24/5 and -3

Using eq  we see that
when x = -24/5, y = -7/5
when x = -3, y = 4

These are the solutions for x and y, and geometrically these coordinates represent the points of intersection of the circle  and the straight line , with reference to the general equation of a circle:

(x - a)^2 + (y - b)^2 = r^2, where r is the radius of the circle, and (a,b) is the centre of the circle

and the general equation for a straight line:

y = mx + c, where m is the gradient of the line and c is the y intercept.

Hope this helps.

Hi - make y=13+3x

then substitute into the first equation - giving you a quadratic in x.

You can then solve that by factorising to get x values and consequently the y values.

(x+3)(5x+24)=0   etc

Graphically this is working out the 2 places that y=3x+3 crosses the circle of radius 5 centre (0,0).

Hope that helps

The best way of solving simultaneous algebraic equation are elimination of variable, matrix methods

we are going to explain elimination method for two equations for two variables as follows:

take two general equations viz

a1x+ b1y=c1 . .. .. . (1)

a2x+b2y=c2 .... . . . (2)

here

x, y are variables, a1, a2, b1, b2, c1, c2 are constants

here we use elimination method it may be applied to any variable x or y .we apply for x

step1:   we make Coficient of both equations equal for this we multiply (1) by a2 and (2) by a1 and get

a1a2x+b1a2y=c1a2  .. .  . ... (3)

a1a2x+b2a1y=c2a1  ...   .    (4)

now subtract (4) from (3) we get

b1a2y-b2a1y=c1a2-c2a1  .    ..    .   (5)

or

(b1a2-b2a1) y=c1a2-c2a1

or y= (c1a2-c2a1) / (b1a2-b2a1)  ....   (6)

put this value of y in (1)  or (2)  we can get value of x called it (7)

then (6) and(7)  are solution of problem

we use this method in our given problem by replacing Coficients of x and y by given Cofficients.

As a check to see if you've calculated the right numbers, it's always a good idea to put the numbers you've found back into the original simultaneous equations to see if the values hold up.

So for x = -3 , y = 4 , substituting this back into a and b gives us:

(-3)^2 + (4)^2 = 9 + 16 = 25

4 - 3(-3) = 4 9 = 13

For x = -4.8 , y = -1.4

(4.8)^2 + (-1.4)^2 = 23.04 + 1.96 = 25

-1.4 - 3 (-4.8) = -1.4 + 14.4 = 13

I'm assuming that those are two simultaneous equations:

x^2 + y^2  = 25 ---------- a

y - 3x = 13 ------------ b

From the second equation (b), we can rearrange the equation to get an equation for y which requires us to add 3x to both sides of the equal sign.
We then get:

y = 13 + 3x --------- c

This is now in a suitable form to allow us to substitute it back into the first original simultaneous equation (a) so that we have an equation just in terms of x so that we can solve for x:

x^2 + (13 + 3x)^2 = 25

The expansion of (13 + 3x)^2 = 9x^2 + 78x + 169

(If you're unsure about how I arrived at this, you can get in touch with me and I'll be more than happy to explain)

We then have:

x^2 + 9x^2 + 78x + 169 = 25

This is in the form of a quadratic equation, so we can shift all the terms to one side of the equation, equating the equation to 0:

10x^2 + 78x + 144 = 0

There are many ways to solve a quadratic equation. The most straight forward way when you have such big numbers would probably be to use the quadratic formula, but you can also try to find suitable factors. I'll use the quadratic formula which is:

x = (-b +- sqrt(b^2 - 4ac))/2a

This can be seen more clearly on the website (https://www.bbc.com/bitesize/guides/z3hb97h/revision/2)

From our quadratic equation, a = 10 , b = 78 , c = 144

Hence;

x = (-(78) +- sqrt ( 78^2 - 4*10*144))/2*10

x = (-78 +- sqrt (6084 - 5760))/20

x = (-78 +- sqrt (324))/20

x = (-78 +- 18)/20

So therefore:

x = (-78+18)/20   OR    x = (-78-18)/20

This gives us values for x as:

x = -3         OR           x = -4.8

This is not the right place to stop because we still need to find the corresponding values of y from equation c.

y = 13 + 3x

When x = -3      AND          When x = -4.8

y = 13 + 3(-3)                                y = 13 + 3(-4.8)

y = 13 - 9                                       y = 13 - 14.4

y = 4                                              y = -1.4

I hope this helped and if you have any questions, you can get in touch with me! 