Simultaneous equations might look scary, but they're actually just puzzles...
Maths615 views·Updated 30 Jul 2026·6 pagesUnderstanding Simultaneous Equations
Charlotte@charlotte_xx
1 / 6
1
of 6
Getting Started with Elimination
The elimination method is your best mate for solving simultaneous equations quickly. You're basically trying to make one of the variables disappear so you can solve for the other one.
Look at this example: when you subtract the second equation from the first, the 3y terms cancel out perfectly, leaving you with 2x = 8. This means x = 4, and you can substitute this back to find y = 2.
Quick Tip: Always check your answer by plugging both values back into the original equations - if they work, you've nailed it!
The key is spotting which variable will eliminate easily, then using that to your advantage.
2
of 6
Mastering the Substitution Back
Once you've found one variable using elimination, substituting back becomes dead simple. Take the first example: after finding x = 3, you just pop it into either original equation to find y.
With 5x + 4y = 43, substitute x = 3 to get 15 + 4y = 43, which gives you y = 7. The solution is (3,7).
Pro Tip: Choose the simpler-looking equation for substitution - it'll save you time and reduce calculation errors!
Practice makes perfect with these substitutions, so don't worry if it feels clunky at first.
3
of 6
When Coefficients Line Up Perfectly
Sometimes the maths gods smile on you, and the coefficients are already set up for easy elimination. In this example, you've got +4y and -4y, which cancel out beautifully when you add the equations.
Adding 7x + 4y = 58 and 5x - 4y = 14 gives you 12x = 72, so x = 6. Substitute back to find y = 4, and you're done!
Remember: Addition and subtraction both work for elimination - choose whichever makes the coefficients cancel out.
This is the dream scenario that makes simultaneous equations feel like a breeze.
4
of 6
Multiplying to Make It Work
When the coefficients don't naturally eliminate, you need to multiply one or both equations to create matching terms. This is where simultaneous equations get a bit more involved, but the process stays the same.
In the first example, multiplying the first equation by 2 and the second by 3 creates matching y coefficients (6y). Then you can subtract to eliminate y and solve for x.
Strategy Alert: Look for the lowest common multiple of the coefficients you want to eliminate - it keeps the numbers manageable.
The second example shows multiplying by larger numbers, but the principle is identical. Find x first, then substitute to get y.
5
of 6
Real-World Problems with Coffee and Cakes
Word problems with simultaneous equations pop up everywhere, especially in shops and pricing scenarios. The trick is translating the words into maths.
Let x = price of coffee and y = price of cake. "2 coffees and 3 cakes cost £9.95" becomes 2x + 3y = 9.95. "1 coffee and 4 cakes cost £10.35" becomes x + 4y = 10.35.
Translation Tip: Always define your variables clearly at the start - it prevents confusion later!
Solve these exactly like any other simultaneous equations, then remember to interpret your answer in context (coffee costs £2, cake costs £1.65).
6
of 6
Sweet Problems with Packs
This sweets problem shows how simultaneous equations handle real inventory situations. Let x = sweets in small pack and y = sweets in big pack.
"4 small packs and 3 big packs contain 175 sweets" gives you 4x + 3y = 175. "5 small packs and 2 big packs contain 154 sweets" gives you 5x + 2y = 154.
Real-World Reminder: Your answers should make sense - negative sweets or fractional packs usually mean you've made an error!
Using elimination (multiply first equation by 2, second by 3), you'll find that small packs contain 16 sweets and big packs contain 37 sweets.
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Maths615 views·Updated 30 Jul 2026·6 pagesUnderstanding Simultaneous Equations
Charlotte@charlotte_xx
Simultaneous equations might look scary, but they're actually just puzzles where you find two unknown numbers that work in both equations. Once you master the elimination method, you'll be solving these confidently in no time!
1
of 6
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Getting Started with Elimination
The elimination method is your best mate for solving simultaneous equations quickly. You're basically trying to make one of the variables disappear so you can solve for the other one.
Look at this example: when you subtract the second equation from the first, the 3y terms cancel out perfectly, leaving you with 2x = 8. This means x = 4, and you can substitute this back to find y = 2.
Quick Tip: Always check your answer by plugging both values back into the original equations - if they work, you've nailed it!
The key is spotting which variable will eliminate easily, then using that to your advantage.
2
of 6Sign up to see the content. It's free!
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By signing up you accept Terms of Service and Privacy Policy
Mastering the Substitution Back
Once you've found one variable using elimination, substituting back becomes dead simple. Take the first example: after finding x = 3, you just pop it into either original equation to find y.
With 5x + 4y = 43, substitute x = 3 to get 15 + 4y = 43, which gives you y = 7. The solution is (3,7).
Pro Tip: Choose the simpler-looking equation for substitution - it'll save you time and reduce calculation errors!
Practice makes perfect with these substitutions, so don't worry if it feels clunky at first.
3
of 6Sign up to see the content. It's free!
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When Coefficients Line Up Perfectly
Sometimes the maths gods smile on you, and the coefficients are already set up for easy elimination. In this example, you've got +4y and -4y, which cancel out beautifully when you add the equations.
Adding 7x + 4y = 58 and 5x - 4y = 14 gives you 12x = 72, so x = 6. Substitute back to find y = 4, and you're done!
Remember: Addition and subtraction both work for elimination - choose whichever makes the coefficients cancel out.
This is the dream scenario that makes simultaneous equations feel like a breeze.
4
of 6Sign up to see the content. It's free!
- Access to all documents
- Improve your grades
- Join milions of students
By signing up you accept Terms of Service and Privacy Policy
Multiplying to Make It Work
When the coefficients don't naturally eliminate, you need to multiply one or both equations to create matching terms. This is where simultaneous equations get a bit more involved, but the process stays the same.
In the first example, multiplying the first equation by 2 and the second by 3 creates matching y coefficients (6y). Then you can subtract to eliminate y and solve for x.
Strategy Alert: Look for the lowest common multiple of the coefficients you want to eliminate - it keeps the numbers manageable.
The second example shows multiplying by larger numbers, but the principle is identical. Find x first, then substitute to get y.
5
of 6Sign up to see the content. It's free!
- Access to all documents
- Improve your grades
- Join milions of students
By signing up you accept Terms of Service and Privacy Policy
Real-World Problems with Coffee and Cakes
Word problems with simultaneous equations pop up everywhere, especially in shops and pricing scenarios. The trick is translating the words into maths.
Let x = price of coffee and y = price of cake. "2 coffees and 3 cakes cost £9.95" becomes 2x + 3y = 9.95. "1 coffee and 4 cakes cost £10.35" becomes x + 4y = 10.35.
Translation Tip: Always define your variables clearly at the start - it prevents confusion later!
Solve these exactly like any other simultaneous equations, then remember to interpret your answer in context (coffee costs £2, cake costs £1.65).
6
of 6Sign up to see the content. It's free!
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Sweet Problems with Packs
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"4 small packs and 3 big packs contain 175 sweets" gives you 4x + 3y = 175. "5 small packs and 2 big packs contain 154 sweets" gives you 5x + 2y = 154.
Real-World Reminder: Your answers should make sense - negative sweets or fractional packs usually mean you've made an error!
Using elimination (multiply first equation by 2, second by 3), you'll find that small packs contain 16 sweets and big packs contain 37 sweets.
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Students love us — and so will you.
4.6/5App Store
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