Topics › KS4 › Algebra · Foundation & Higher

Simultaneous equations

1. sim-01 Present from here

   Solve the simultaneous equations $2x + y = 13$ and $x + y = 8$

   A Option A: $x = 5, y = 3$

   B Option B: $x = 3, y = 5$

   C Option C: $x = 7, y = 1$

   D Option D: $x = 5, y = 13$
   Answer and misconceptions

   A Option A: $x = 5, y = 3$ ✓ Correct

   B Option B: $x = 3, y = 5$ ✗

   Swaps the values of $x$ and $y$Finds $5$ and $3$ but gives them to the wrong letters. Check: $2 \times 3 + 5 = 11$, not $13$.

   C Option C: $x = 7, y = 1$ ✗

   Adds the equations when the $y$ terms have the same signAdding gives $3x + 2y = 21$, so $y$ is not eliminated. Both $y$ terms are $+y$, so subtract the equations.

   D Option D: $x = 5, y = 13$ ✗

   Adds instead of subtracting when finding $y$From $5 + y = 8$, works out $y = 8 + 5$; subtract $5$ from both sides to get $y = 3$.
2. sim-02 Present from here

   $y = 2x$ and $x + y = 12$. What is the value of $x$?

   A Option A: $6$

   B Option B: $5$

   C Option C: $4$

   D Option D: $8$
   Answer and misconceptions

   A Option A: $6$ ✗

   Thinks $x + 2x = 2x$After substituting, $x + 2x = 3x$, so $3x = 12$.

   B Option B: $5$ ✗

   Treats $2x$ as $x + 2$Writes $x + x + 2 = 12$; $2x$ means $2 \times x$, not $x + 2$.

   C Option C: $4$ ✓ Correct

   D Option D: $8$ ✗

   Gives the value of $y$ instead of $x$Correctly finds $x = 4$ and $y = 8$, but the question asks for $x$.
3. sim-03 Present from here

   $5x + 2y = 20$ and $3x + 2y = 16$. What is the value of $y$?

   A Option A: $10$

   B Option B: $5$

   C Option C: $11$

   D Option D: $7$
   Answer and misconceptions

   A Option A: $10$ ✗

   Stops at $2y$ instead of finding $y$Gets $2y = 10$ but does not divide by $2$.

   B Option B: $5$ ✓ Correct

   C Option C: $11$ ✗

   Adds instead of subtracting when rearrangingFrom $6 + 2y = 16$, works out $2y = 16 + 6$; subtract $6$ from both sides: $2y = 10$.

   D Option D: $7$ ✗

   Replaces $3x$ with the value of $x$Substitutes $x = 2$ as $2 + 2y = 16$; $3x$ means $3 \times 2 = 6$.
4. sim-04 Present from here

   Two adult tickets and one child ticket cost £26. One adult ticket and one child ticket cost £17. How much is one child ticket?

   A Option A: £9

   B Option B: £8.50

   C Option C: £4

   D Option D: £8
   Answer and misconceptions

   A Option A: £9 ✗

   Thinks subtracting the equations gives the child price$26 - 17 = 9$ is the cost of the extra adult ticket, so £9 is the adult price; the child ticket is £17 − £9.

   B Option B: £8.50 ✗

   Assumes adult and child tickets cost the sameWorks out £17 ÷ 2; the two prices are different, so use both facts.

   C Option C: £4 ✗

   Ignores the child ticket in the first statementWorks out £26 ÷ 2 = £13 for an adult, but the £26 also includes a child ticket.

   D Option D: £8 ✓ Correct
5. sim-05 Present from here

   Which pair of values satisfies both $x + y = 9$ and $2x - y = 3$?

   A Option A: $x = 5, y = 4$

   B Option B: $x = 12, y = -3$

   C Option C: $x = 4, y = 5$

   D Option D: $x = 4, y = -5$
   Answer and misconceptions

   A Option A: $x = 5, y = 4$ ✗

   Swaps the values of $x$ and $y$Gives $4$ and $5$ to the wrong letters. Check: $2 \times 5 - 4 = 6$, not $3$.

   B Option B: $x = 12, y = -3$ ✗

   Does not multiply every term in the bracket by the minus signSubstitutes $y = 9 - x$ and writes $2x - (9 - x)$ as $2x - 9 - x$. The minus sign applies to both terms: $2x - 9 + x = 3$, so $3x = 12$.

   C Option C: $x = 4, y = 5$ ✓ Correct

   D Option D: $x = 4, y = -5$ ✗

   Loses a negative sign when solving for $y$From $-y = -5$, writes $y = -5$; dividing both sides by $-1$ gives $y = 5$.