/* Base Styles (Mobile First) */
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font-family: ‘Inter’, Arial, sans-serif;
color: #515151;
max-width: 1300px;
margin: 0 auto;
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.r-section {
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font-weight: 700;
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}
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max-width: 800px;
margin: 0 auto 30px auto;
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text-align: center;
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max-width: 800px;
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/* Desktop Styles */
@media (min-width: 768px) {
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Solving Simultaneous Linear Equations
Simultaneous equations help us solve real-world puzzles with multiple unknown values, like working out the individual cost of items in a meal deal. This guide will show you the graphical, elimination, and substitution methods to find the values that make both equations true at the same time.
Three Methods to Solve Simultaneous Equations
1. Graphical Method
Plot both linear equations on the same graph. The point where the two lines intersect (cross) is the solution. This method is great for visualising the answer but often gives an approximate solution.
2. Elimination Method
Add or subtract the two equations to eliminate one of the variables. This leaves you with a simple one-variable equation to solve. This method is best when the coefficients of one variable are the same or opposites.
3. Substitution Method
Rearrange one equation to isolate a variable (e.g., get $x$ on its own). Then, substitute this expression into the other equation. This also leaves you with a one-variable equation to solve.
Worked Examples
Example 1: Elimination Method
Solve: $3x + 2y = 13$ and $2x – 3y = 0$.
- Make the y-coefficients match. Multiply the first equation by 3 and the second by 2 to get:
$9x + 6y = 39$
$4x – 6y = 0$ - The signs of $y$ are different, so add the equations to eliminate $y$.
$(9x+4x) + (6y-6y) = 39+0 \implies 13x = 39$. - Solve for $x$: $x=3$.
- Substitute $x=3$ into an original equation: $2(3) – 3y = 0 \implies 6 = 3y \implies y=2$.
Answer: $x=3, y=2$
Example 2: Substitution Method
Solve: $y = x + 2$ and $2x + y = 11$.
- The first equation already has $y$ isolated.
- Substitute $(x+2)$ for $y$ in the second equation:
$2x + (x+2) = 11$. - Solve the new equation for $x$:
$3x + 2 = 11 \implies 3x = 9 \implies x=3$. - Substitute $x=3$ back into the first equation:
$y = 3 + 2 = 5$.
Answer: $x=3, y=5$
Tutor Insights
🤔 Common Misunderstandings
- Elimination: Not knowing whether to add or subtract the equations. Remember: Same Signs Subtract, Opposite Signs Add.
- Substitution: Substituting an expression back into the same equation it came from, which leads to a dead end like $0=0$.
📝 Common Exam Mistakes
- Sign errors: Especially when subtracting an entire equation or multiplying by a negative number.
- Only finding one variable: Solving for $x$ and then forgetting to substitute back to find $y$.
- Not showing working: You must show your algebraic steps to get full marks.
Practice Questions
- Solve using elimination: $x + y = 9$ and $x – y = 3$.
- Solve using substitution: $y = x + 2$ and $2x + y = 11$.
- Solve: $4x + 2y = 16$ and $x + 2y = 7$.
- A cinema sells adult tickets for £8 and child tickets for £5. 150 tickets were sold for a total of £975. How many of each ticket were sold?
Show Answers
- Working: Add the equations: $(x+x) + (y-y) = 9+3 \implies 2x=12 \implies x=6$. Substitute back: $6+y=9 \implies y=3$.
Answer: $x=6, y=3$. - Working: Substitute $(x+2)$ into the second equation: $2x+(x+2)=11 \implies 3x=9 \implies x=3$. Substitute back: $y=3+2=5$.
Answer: $x=3, y=5$. - Working: Subtract the second equation from the first: $(4x-x) + (2y-2y) = 16-7 \implies 3x=9 \implies x=3$. Substitute back: $3+2y=7 \implies 2y=4 \implies y=2$.
Answer: $x=3, y=2$. - Working: Let $a$ be adult tickets and $c$ be child tickets. Equations are: $a+c=150$ and $8a+5c=975$. Rearrange first: $a=150-c$. Substitute into second: $8(150-c)+5c=975 \implies 1200-8c+5c=975 \implies 1200-3c=975 \implies 225=3c \implies c=75$. Substitute back: $a=150-75=75$.
Answer: 75 adult tickets and 75 child tickets.
