Solve algebraically the simultaneous equations y-3x=13 and x^2+y^2=25

The point of solving simultaneous equations is to find the points at which the two equations intersect. In this case, it is clear to see that the first one is a circle with radius 5, centre (0,0) and the other is a line. When you’re given a set of equations, there are usually two ways to go about solving the problem. The first way is elimination where you can add or take away multiples of one equation from the other so that you end up with an equation consisting of one variable. This means that you can substitute your answer into either equation to get the other variable, and that is your solution. However, the second way substitution, is more applicable to this question since there are powers of x and y meaning you can’t add or take away multiples from the other.For this question, you can easily rearrange y-3x=13 either to y=13+3x or x=(y-13)/3. It’s usually easier and simpler to work without fractions, so let’s choose the former. Now that you have an equation y in terms of x, we can substitute into x^2+y^2=25, which gives x^2+(13+3x)^2=25. Now to solve this, we have to expand the brackets and simplify. So, the last equation is the same as writing x^2+(13+3x)(13+3x)=25. This can be expanded Yeah to x^2+169+39x+39x+9x^2=25, and simplifying gives 10x^2+78x+144=0 (). This can be easily solved using the quadratic equation, which is x=[-b+/-(b^2-4ac)^(1/2)]/2a, which solves an equation of the form ax^2+bx+c=0. Applying this to our equation (), we get that x=-3 and x=-24/5. Now using these solutions, you can substitute into y-3x=13 to find y. So, for x=-3, we have y=4 and for x=-24/5, we have y=-7/5. So the points of intersection are (-3,4) and (-24/5,-7/5). Notice that there are two solutions as the intersection of a circle with a line gives two points of intersection.