If one equation is linear and the other involves x^2 then you should use elimination.
If you have y=mx+c then substitute (mx+c) into your second equation instead of y. Then rearrange it and solve.
A linear equation is of the form y = mx+c (it's a straight line)
A quadratic equation is of the form y = ax2+bx+c, the highest power is x2 (it's a curve)
Graphically, the solutions for x and y are given by the coordinates at which the straight line cuts the x2 graph.
There may be 2 solutions for x and y, only 1 solution, or no solutions.
In the case where there is 1 solution, the linear graph will be a tangent to the quadratic at some point.
In the case where there are no solutions, the linear graph will not touch or cut the quadratic.
Method of Substitution
If it is not possible to draw the simualtaneous linear equations accurately, we can use algebra in the method of substitution.
1) Rearrange the linear equation into terms of x or y. So y = .. or x = ..
2) Substitute this expression into the quadratic equation, so that you now have an equation in terms of one variable, and you can solve it.
3) Plug your answer back into one of the equation, and solve for the other variable.
4) Check your solution with the other equation.
a) y = 2x + 1
b) y = x2 + 1
1) They are both in terms of y = .. so we don't need to do any rearranging.
2) We will substitute a) into b) to get:
2x + 1 = x2 + 1
We now rearrange to solve for x:
x2 + 1 - 1 - 2x = 0
x2 - 2x = 0
x (x - 2) = 0
So x = 0 or x =2.
3) We have two answers for x so we will have 2 answers, and we need to consider them seperately.
If x = 0:
a) y = 2(0) + 1 = 1
4) Let's check the solution [x = 0, y = 1] works for b)
x2 + 1 = 1 = y
It works so [x = 0, y = 1] is one solution.
Let's work out the other.
For x =2, we subsitute it in to a):
y = 2(2) + 1 = 5
lets check [x=2, y=5] works with equation b):
y = 22+1 = 5.
So our two solutions are (0,1) and (2,5)
You can check this also by drawing it!
Nothing in this section yet. Why not help us get started?