Coaching Centre
Simultaneous equations algebraically
Solving simultaneous linear equations exactly by substitution and by elimination, and setting up a pair of equations from a worded problem.
A graph shows the idea, but it is slow and often only gives an estimate. Algebra gives the exact answer every time. The plan for both methods is the same: get rid of one letter, so you are left with an ordinary one-variable equation you already know how to solve.
Substitution
Use substitution when one equation already says "$y = \ldots$" or "$x = \ldots$". Because $y$ is equal to that expression, you can swap the expression in wherever $y$ appears in the other equation.
Solve $y = 3x - 4$ and $2x + y = 11$.
Elimination
Use elimination when both equations look like $ax + by = c$. If the $x$ numbers (or the $y$ numbers) are opposites, adding the equations makes that letter disappear. If they are equal, subtract instead. If they are neither, multiply one or both equations first to make them match.
Solve $2x + 3y = 12$ and $5x - 2y = 11$.
On Pair 1, find multipliers that make the $y$ terms opposites, then find a different pair that eliminates $x$ instead. Do both routes give the same solution? On Tickets, the first equation counts people and the second counts dollars: what multiplier on the first equation eliminates $x$ in one step?
From words to equations
Most exam questions are in words. Name the two unknowns, write one equation for each fact, then solve. Two unknowns always need two separate facts.
A school musical sold $230$ tickets and took $\$4530$. Adult tickets cost $\$24$ and student tickets $\$15$. How many of each were sold?
A café sells a coffee and two muffins for $\$13.50$, and three coffees and one muffin for $\$17$. Find the price of each.
Your turn
Solve $3x + 4y = 5$ and $2x - 3y = 9$ by elimination.
Show the working
Multiply the first by $2$ and the second by $3$: $6x + 8y = 10$ and $6x - 9y = 27$. Subtract: $17y = -17$, so $y = -1$. Then $3x - 4 = 5$, so $x = 3$. Check: $2 \times 3 - 3 \times (-1) = 9$ ✓
Solve $y = 3x - 4$ and $2x + y = 11$. What is $x$?
$2x + 3x - 4 = 11$, so $5x = 15$ and $x = 3$ (then $y = 5$).
Solve $2x + 3y = 12$ and $5x - 2y = 11$. What is $y$?
Multiply by $2$ and $3$: $4x + 6y = 24$ and $15x - 6y = 33$. Add: $19x = 57$, $x = 3$. Then $6 + 3y = 12$, so $y = 2$.
To solve $x + y = 10$ and $x - y = 4$, the quickest first step is to:
Adding gives $2x = 14$, so $x = 7$ and $y = 3$, with no multiplying needed.
Solve $x + y = 10$ and $x - y = 4$. What is $x$?
Add: $2x = 14$, so $x = 7$ (and $y = 3$).
What do you get when you multiply $3x - 2y = 7$ by $4$?
Every term is multiplied: $4 \times 3x = 12x$, $4 \times (-2y) = -8y$, $4 \times 7 = 28$.
Solve $3x + 4y = 5$ and $2x - 3y = 9$. Give x, y.
$6x + 8y = 10$ and $6x - 9y = 27$. Subtract: $17y = -17$, $y = -1$. Then $3x - 4 = 5$, $x = 3$.
One coffee and two muffins cost $\$13.50$. Three coffees and one muffin cost $\$17$. What does one coffee cost, in dollars?
$m = 17 - 3c$. Then $c + 2(17 - 3c) = 13.5$, so $-5c = -20.5$ and $c = 4.1$.
A word problem gives $a = 12.4$ buses. What should you do?
A non-whole count means an error or impossible data. Find the problem before giving an answer.
Exam style: A concert sold $300$ tickets for a total of $\$6030$. Adult tickets cost $\$24$ and student tickets $\$15$. How many adult tickets were sold?
$a + s = 300$ and $24a + 15s = 6030$. Multiply the first by $15$: $15a + 15s = 4500$. Subtract: $9a = 1530$, so $a = 170$.
Exam style: How many student tickets were sold at that concert?
$s = 300 - 170 = 130$. Check: $24 \times 170 + 15 \times 130 = 4080 + 1950 = 6030$ ✓