UK Secondary (7-11)
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Systems of linear and quadratic equations

Interactive practice questions

Consider the following system of equations:

$x^2$x2 $=$= $y+14$y+14
$y$y $=$= $3x-16$3x16

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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=3x16

           
$LHS$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=3x16

           
$LHS$LHS $=$= $\editable{}$   $RHS$RHS $=$= $3\times\editable{}-16$3×16
          $=$= $\editable{}$
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When solving $y=x^2-3x+1$y=x23x+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?

Where does the vertical line $x=-5$x=5 intersect the curve $y=-2x^2+x-12$y=2x2+x12?

Solve the following equations.

Equation 1 $y=x^2$y=x2
Equation 2 $y=9$y=9

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