Year 6

Finding solutions to problems with two variables

Year 6

Finding solutions to problems with two variables

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Lesson details

Key learning points

  1. In this lesson, we will be discussing how to work systematically to find all possibilities for problems with two variables.

Licence

This content is made available by Oak National Academy Limited and its partners and licensed under Oak’s terms & conditions (Collection 1), except where otherwise stated.

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4 Questions

Q1.
Which expression would represent the following problem: The cooking time for roast beef is 30 minutes for every kg plus an additional 15 minutes.
Correct answer: a = 30b + 15
a = 500b + 30
Q2.
Which expression would represent the following problem: The cost of a photo album book is £2.50 for binding and printing with an additional cost of 15p per photograph.
Correct answer: a = 15b +250
a = 250b + 15
Q3.
Which expression would represent the following problem: A singer makes money from selling music. They pay a one-off fee of £25.00 to Oak-Tunes for marketing and distribution costs and receive 55p for each copy of the album sold.
a = 25 + 0.55
a = 2500 + 55a
Correct answer: a = 55b - 2500
Q4.
Instead of writing 5 x a in an algebraic expression, what could I write instead?
5 + a
Correct answer: 5a
a5

4 Questions

Q1.
3f + g = 15. Which of the following pairs of numbers would work?
f = 2, g = 10
Correct answer: f = 3, g = 6
f = 4, g = 2
f = 5, g = 1
Q2.
5a - 2b = 8. Which of the following pairs of numbers would work?
a = 3, b = 2
a = 4, b = 12
Correct answer: a = 4, b = 6
a = 5, b = 6
Q3.
3c + 2d = 20. Which pair of numbers does does NOT work with this equation?
c = 2, d = 7
Correct answer: c = 5, d = 2
c = 6, d = 1
Q4.
6m - 3n = 15. Which pair of numbers does does NOT work with this equation?
m = 3, n = 1
Correct answer: m = 4, n = 2
m = 5, n = 5
m = 7, n = 9