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.

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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