Coaching Centre
Simultaneous equations
Solve pairs of equations by substitution and elimination, including a line meeting a parabola, and see each solution as an intersection point.
What a solution means
A solution of a pair of equations is a pair $(x, y)$ that makes both equations true at once. On a graph, it is a point where the two graphs meet.
Elimination
Multiply one or both equations so a variable has matching coefficients, then add or subtract to eliminate it.
Solve $2x + 3y = 12$ and $5x - 2y = 11$.
Substitution: a line and a parabola
When one equation is quadratic, substitute the linear one into it. You get a quadratic equation, which can have $2$, $1$ or $0$ solutions.
Solve $y = x^2 - 3$ and $y = 2x$.
- Slide $b$ down until the line just touches the parabola. That line is a tangent.
- Find a line that misses the parabola completely.
Solve $x + y = 7$ and $x - y = 1$.
Add: $2x = 8$, so $x = 4$ and $y = 3$.
How many solutions do $y = x^2$ and $y = -1$ have?
$x^2 = -1$ has no real solutions: the line is below the parabola.
The line $y = 2x - 1$ and the parabola $y = x^2$ meet at exactly one point. Find its $x$-coordinate.
$x^2 - 2x + 1 = 0$, so $(x - 1)^2 = 0$ and $x = 1$.
Solve $y = 3x - 1$ and $y = x + 5$. Enter $x$.
$3x - 1 = x + 5$, so $2x = 6$ and $x = 3$ (then $y = 8$).
Solve $3x + y = 11$ and $x - y = 1$.
Add: $4x = 12$, so $x = 3$ and $y = 2$.
Solve $y = x^2$ and $y = x + 2$. Enter both $x$-values.
$x^2 - x - 2 = 0$, so $(x - 2)(x + 1) = 0$.
A line and a parabola can meet in at most:
Substituting gives a quadratic, which has at most two solutions.
Solve $2x + 3y = 13$ and $4x - 3y = -1$. Enter $y$.
Add: $6x = 12$, so $x = 2$. Then $4 + 3y = 13$ gives $y = 3$.
New numbers every time. Keep going until you get five in a row.