In an earlier Section you solved simultaneous equations where both of the equations were linear. In this section we extend this to solving simultaneous equations where one equation is linear and the other is quadratic. This will normally give you a quadratic equation to solve.
Worked Examples
Solve the simultaneous equations
| y | = x2 − 1 | (1) |
| y | = 5 − x | (2) |
Substituting y from equation (1) into equation (2) gives
x2 − 1 = 5 − x
This equation simplifies to
0 = x2 − 1 − 5 + x
so0 = x2 + x − 6
We now solve this quadratic equation by factorisation.
0 = (x + 3)(x − 2)
sox + 3 = 0 or x − 2 = 0
thereforex = −3 or x = 2
These values of x are now substituted into one of the original equations to find the corresponding values of y. It is usually easiest to use the linear equation for this.
Substituting the first solution x = −3 into equation (2) gives y = 5 −(−3) = 8.
Substituting the second solution x = 2 into equation (2) gives y = 5 − 2 = 3.
The solutions are x = −3, y = 8 and x = 2, y = 3.
Note
In Worked Example 1, the first equation, y = x2 − 1, can be represented by a quadratic curve graph. The second equation, y = 5 − x , can be represented by a straight line. When we solve this pair of simultaneous equations, we are finding the coordinates of the two points where the line and the curve intersect. This is shown on the graph, the intersection points being (–3, 8) and (2, 3).
Solve the simultaneous equations
| y | = 3x2 − 4 | (1) |
| y | = 2x + 3 | (2) |
Subtract equation (2) from equation (1).
| y | = 3x2 − 4 | (1) |
| y | = 2x + 3 | (2) |
| 0 | = (3x2 − 4) − (2x + 3) | (2) − (1) |
This equation simplifies to
0 = 3x2 − 4 − 2x − 3
0 = 3x2 − 2x − 7
This quadratic does not factorise, so we need to use the quadratic formula
x =
to find the two possible values of x. Here
a = 3, b = −2 and c = −7
So
x = = =
| so | x | = 1.896805253 |
| = 1.90 (to 2 d.p.) | ||
| or | x | = −1.230138587 |
| = −1.23 (to 2 d.p.) |
Substituting the un-rounded values for x into the linear equation (2) gives
y = 2x + 3 = 2 × 1.896805253 + 3 = 6.793610507 = 6.79 (to 2 d.p.)
and
y = 2x + 3 = 2 × −1.230138587 + 3 = 0.539722826 = 0.54 (to 2 d.p.).
The solutions are x = 1.90, y = 6.79 and x = −1.23, y = 0.54 (to 2 d.p.).
Solve the simultaneous equations
| y | = x2 + 2x − 3 | (1) |
| y | = 2x | (2) |
Subtract equation (2) from equation (1).
| y | = x2 + 2x − 3 | (1) |
| y | = 2x | (2) |
| 0 | = (x2 + 2x − 3) − (2x) | (2) − (1) |
This equation simplifies to
| 0 = x2 − 3 | |
| so | x2 = 3 |
| hence | x = ± |
These values for x are now substituted into the linear equation.
Substituting the first solution, x = , into equation (2) gives y = 2.
Substituting the second solution, x = −, into equation (2) gives y = 2(−) = −2.
The solutions are x = , y = 2 and x = −, y = −2.
Solve algebraically the simultaneous equations
| x2 + y2 | = 10 | (1) |
| y | = x + 2 | (2) |
We use the linear equation (2) to substitute for y2 into equation (1), the equation for the circle.
The square of equation (2) is
y2 = (x + 2)2
soy2 = x2 + 4x + 4
Substituting into equation (1) now gives
x2 + (x2 + 4x + 4) = 10
which simplifies to2x2 + 4x − 6 = 0
which, dividing by 2, simplifies again to
x2 + 2x − 3 = 0
We can now solve this equation for x by factorising
(x + 3)(x − 1) = 0
sox = −3 or x = 1
When x = −3, equation (2) gives
y = −3 + 2 = −1
When x = 1, equation (2) gives
y = 1 + 2 = 3
Therefore the solutions of the two equations are
x = −3, y = −1 and x = 1, y = 3
Write down the coordinates of the points of intersection of the circle x2 + y2 = 10 and the straight line y = x + 2.
Show meThe points of intersection of the circle x2 + y2 = 10 and the straight line y = x + 2 are (−3, −1) and (1, 3).
Exercises
Where necessary round answers to 2 decimal places.