# A19a – Solving two linear simultaneous equations algebraically by elimination

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A starting point...

If a cake weighs $$600$$ g and a muffin weighs $$120$$ g, what is the mass of 3 cakes and 2 muffins?

The answer is $$3 \times 600 + 2 \times 120 = 2040$$ g

What is the mass of 6 cakes and 4 muffins?

The slow way to work this out is to do $$6 \times 600 + 4 \times 120 = 4080$$ g.

A quicker way is to spot that 6 cakes and 4 muffins must weigh twice as much as 3 cakes and 2 muffins, and $$2 \times 2040 = 4080$$ g.

Now suppose we know that 3 cakes and 2 muffins weigh 2040 g, but that we don’t know how much an individual cake or muffin weighs. Consider the following questions:

Do you have enough information to work out how the mass of an individual cake or an individual muffin?

Do you have enough information to work out the total mass of 6 cakes and 4 muffins?

Do you have enough information to work out the total mass of 42 cakes and 28 muffins?

Do you have enough information to work out the total mass of 6 cakes and 6 muffins?

# Part 1 – Solving a pair of linear simultaneous equations by elimination (easy)

Part 1 contains easy examples where one variable either has the same coefficient in the two equations, or it has the same coefficient with the opposite sign.

Teacher resources for Part 1

# Part 2 – Solving a pair of simultaneous equations by elimination (hard)

Part 2 contains harder examples where students need to multiply one or both equations by suitable integers before they can eliminate a variable.

Teacher resources for Part 2