Quadratic Equations
Hong Kong
Stage 4 - Stage 5
Simultaneous, linear and quadratic equations
Interactive practice questions
Consider the following system of equations:
| $x^2$x2 | $=$= | $y+14$y+14 |
| $y$y | $=$= | $3x-16$3x−16 |
Loading Graph...
a
By considering the left hand side ($LHS$LHS) and right hand side ($RHS$RHS) of each equation, fill in the missing values to verify that the points of intersection on the graphs are solutions of the corresponding system of equations.
First, test the point $\left(1,-13\right)$(1,−13).
|
$x^2=y+14$x2=y+14 |
||||||
|---|---|---|---|---|---|---|
| $LHS$LHS | $=$= | $\left(\editable{}\right)^2$()2 | $RHS$RHS | $=$= | $\editable{}+14$+14 | |
| $=$= | $\editable{}$ | $=$= | $\editable{}$ | |||
|
$y=3x-16$y=3x−16 |
||||||
| $LHS$LHS | $=$= | $\editable{}$ | $RHS$RHS | $=$= | $3\times\editable{}-16$3×−16 | |
| $=$= | $\editable{}$ |
b
Now test the point $\left(2,-10\right)$(2,−10).
|
$x^2=y+14$x2=y+14 |
||||||
|---|---|---|---|---|---|---|
| $LHS$LHS | $=$= | $\left(\editable{}\right)^2$()2 | $RHS$RHS | $=$= | $\editable{}+14$+14 | |
| $=$= | $\editable{}$ | $=$= | $\editable{}$ | |||
|
$y=3x-16$y=3x−16 |
||||||
| $LHS$LHS | $=$= | $\editable{}$ | $RHS$RHS | $=$= | $3\times\editable{}-16$3×−16 | |
| $=$= | $\editable{}$ |
Easy
3min
When solving $y=x^2-3x+1$y=x2−3x+1 and $y=x+6$y=x+6 simultaneously, one point of intersection is at $x=-1$x=−1. What is the $y$y-coordinate at this point?
Easy
1min
Where does the vertical line $x=-5$x=−5 intersect the curve $y=-2x^2+x-12$y=−2x2+x−12?
Easy
1min
Solve the following equations.
| Equation 1 | $y=x^2$y=x2 |
| Equation 2 | $y=9$y=9 |
Easy
1min
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