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MathsMaths599 views·Updated May 28, 2026·6 pages

Understanding Simultaneous Equations

user profile picture
Charlotte@charlotte_xx

Simultaneous equations might look scary, but they're actually just puzzles... Show more

1
of 6
Simultaneous Equatias
-
$5x + 3y = 26$
$3x + 3y = 18$
when x = 4
$2x = 8$
Sx4 + 3y = 26
$20+ 3y = 26$
(4,2)
$x = 4$
$3y = 6$
$y = 2$

$6x +

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
Simultaneous Equatias
-
$5x + 3y = 26$
$3x + 3y = 18$
when x = 4
$2x = 8$
Sx4 + 3y = 26
$20+ 3y = 26$
(4,2)
$x = 4$
$3y = 6$
$y = 2$

$6x +

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
Simultaneous Equatias
-
$5x + 3y = 26$
$3x + 3y = 18$
when x = 4
$2x = 8$
Sx4 + 3y = 26
$20+ 3y = 26$
(4,2)
$x = 4$
$3y = 6$
$y = 2$

$6x +

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
Simultaneous Equatias
-
$5x + 3y = 26$
$3x + 3y = 18$
when x = 4
$2x = 8$
Sx4 + 3y = 26
$20+ 3y = 26$
(4,2)
$x = 4$
$3y = 6$
$y = 2$

$6x +

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
Simultaneous Equatias
-
$5x + 3y = 26$
$3x + 3y = 18$
when x = 4
$2x = 8$
Sx4 + 3y = 26
$20+ 3y = 26$
(4,2)
$x = 4$
$3y = 6$
$y = 2$

$6x +

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
Simultaneous Equatias
-
$5x + 3y = 26$
$3x + 3y = 18$
when x = 4
$2x = 8$
Sx4 + 3y = 26
$20+ 3y = 26$
(4,2)
$x = 4$
$3y = 6$
$y = 2$

$6x +

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.

We thought you’d never ask...

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MathsMaths599 views·Updated May 28, 2026·6 pages

Understanding Simultaneous Equations

user profile picture
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
Simultaneous Equatias
-
$5x + 3y = 26$
$3x + 3y = 18$
when x = 4
$2x = 8$
Sx4 + 3y = 26
$20+ 3y = 26$
(4,2)
$x = 4$
$3y = 6$
$y = 2$

$6x +

Sign up to see the content. It's free!

  • Access to all documents
  • Improve your grades
  • Join milions of students

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
Simultaneous Equatias
-
$5x + 3y = 26$
$3x + 3y = 18$
when x = 4
$2x = 8$
Sx4 + 3y = 26
$20+ 3y = 26$
(4,2)
$x = 4$
$3y = 6$
$y = 2$

$6x +

Sign up to see the content. It's free!

  • Access to all documents
  • Improve your grades
  • Join milions of students

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
Simultaneous Equatias
-
$5x + 3y = 26$
$3x + 3y = 18$
when x = 4
$2x = 8$
Sx4 + 3y = 26
$20+ 3y = 26$
(4,2)
$x = 4$
$3y = 6$
$y = 2$

$6x +

Sign up to see the content. It's free!

  • Access to all documents
  • Improve your grades
  • Join milions of students

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
Simultaneous Equatias
-
$5x + 3y = 26$
$3x + 3y = 18$
when x = 4
$2x = 8$
Sx4 + 3y = 26
$20+ 3y = 26$
(4,2)
$x = 4$
$3y = 6$
$y = 2$

$6x +

Sign up to see the content. It's free!

  • Access to all documents
  • Improve your grades
  • Join milions of students

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
Simultaneous Equatias
-
$5x + 3y = 26$
$3x + 3y = 18$
when x = 4
$2x = 8$
Sx4 + 3y = 26
$20+ 3y = 26$
(4,2)
$x = 4$
$3y = 6$
$y = 2$

$6x +

Sign up to see the content. It's free!

  • Access to all documents
  • Improve your grades
  • Join milions of students

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
Simultaneous Equatias
-
$5x + 3y = 26$
$3x + 3y = 18$
when x = 4
$2x = 8$
Sx4 + 3y = 26
$20+ 3y = 26$
(4,2)
$x = 4$
$3y = 6$
$y = 2$

$6x +

Sign up to see the content. It's free!

  • Access to all documents
  • Improve your grades
  • Join milions of students

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.

We thought you’d never ask...

What is the Knowunity AI companion?

Our AI Companion is a student-focused AI tool that offers more than just answers. Built on millions of Knowunity resources, it provides relevant information, personalised study plans, quizzes, and content directly in the chat, adapting to your individual learning journey.

Where can I download the Knowunity app?

You can download the app from Google Play Store and Apple App Store.

Is Knowunity really free of charge?

That's right! Enjoy free access to study content, connect with fellow students, and get instant help – all at your fingertips.

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918,818392

Can't find what you're looking for? Explore other subjects.

Students love us — and so will you.

4.6/5App Store
4.7/5Google Play

The app is very easy to use and well designed. I have found everything I was looking for so far and have been able to learn a lot from the presentations! I will definitely use the app for a class assignment! And of course it also helps a lot as an inspiration.

Stefan SiOS user

This app is really great. There are so many study notes and help [...]. My problem subject is French, for example, and the app has so many options for help. Thanks to this app, I have improved my French. I would recommend it to anyone.

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Wow, I am really amazed. I just tried the app because I've seen it advertised many times and was absolutely stunned. This app is THE HELP you want for school and above all, it offers so many things, such as workouts and fact sheets, which have been VERY helpful to me personally.

AnnaiOS user