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Year 7 Maths Algebra and Equations

Substitutions

20 practice questions 1 video lesson Theory + worked examples
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Theory

To substitute, replace each pronumeral with its number, then work out the value using the order of operations. This Year 7 maths page shows how to substitute whole numbers, negative numbers and powers into expressions, how to use a formula, and how substitution checks an expansion.

A pronumeral can stand for different numbers. The expression \(x + 4\) is \(6\) when \(x = 2\), \(11\) when \(x = 7\) and \(24\) when \(x = 20\). Putting a number in place of a pronumeral is called substitution. The answer is the value of the expression.

Step 1. Write the multiplication signs back in: \(5a = 5 \times a\), \(ab = a \times b\), \(a^2 = a \times a\).

Step 2. Replace each pronumeral with its number. A negative number goes in brackets: \(3 \times (-2)\).

Step 3. Calculate in the order of operations: brackets, then powers, then \(\times\) and \(\div\), then \(+\) and \(-\). A fraction line groups the top, so work out the top first, then divide.

A formula is a rule that links pronumerals, such as \(P = 2l + 2w\). State the formula, substitute, calculate, then give the units.

Rectangle 6 cm by 4 cm A rectangle with length 6 centimetres and width 4 centimetres. 6 cm 4 cm 6 cm 4 cm
\(P = 2l + 2w = 2 \times 6 + 2 \times 4 = 12 + 8 = 20\) cm. The formula gives the same perimeter as adding the sides: \(6 + 4 + 6 + 4 = 20\).

Substituting makes a number pattern:

\(n\)\(1\)\(2\)\(3\)\(4\)\(5\)
\(2n + 1\)\(3\)\(5\)\(7\)\(9\)\(11\)
\[ A = \dfrac{1}{2}bh = \dfrac{1}{2} \times 6 \times 7 = 3 \times 7 = 21 \]
A =1/2×6×7=3×7=21
Fractions in a formula. Divide first when you can: \(\dfrac{1}{2} \times 6 \times 7 = 3 \times 7\).
Square a negative in brackets. When \(a = -3\), \(a^2 = (-3)^2 = (-3) \times (-3) = 9\).

How to substitute

  1. Write the multiplication signs back in.
  2. Replace each pronumeral with its number (negatives in brackets).
  3. Calculate: brackets, powers, \(\times\) and \(\div\), then \(+\) and \(-\).
  4. Write the answer, with units for a formula.
Example 1 — Whole numbers
(i) Find the value of \(6x - 5\) when \(x = 3\).
(ii) Complete the table for \(2x + 3\).
\(x\)\(1\)\(2\)\(5\)
\(2x + 3\)???
Solution
\(6 \times 3 - 5\)\(=\)\(18 - 5 = 13\)
\(2 \times 1 + 3\)\(=\)\(5\)
\(2 \times 2 + 3\)\(=\)\(7\)
\(2 \times 5 + 3\)\(=\)\(13\)
6×3−5=13,2×1+3=5,2×2+3=7,2×5+3=13

(i) \(13\)
(ii) \(5, 7, 13\)

Example 2 — Powers
(i) Evaluate \(3m + 5\) when \(m = 10\).
(ii) If \(x = 4y^2 + 5y\), evaluate \(x\) when \(y = 2\).
Solution
\(3 \times 10 + 5\)\(=\)\(30 + 5 = 35\)
\(4 \times 2^2 + 5 \times 2\)\(=\)\(16 + 10 = 26\)
3×10+5=35,4×2+5×2=16+10=26

(i) \(35\)
(ii) \(26\). Square first: \(2^2 = 4\), then \(4 \times 4 = 16\).

Example 3 — Negative numbers
(i) Find \(3p - 2q\) when \(p = 4\) and \(q = -5\).
(ii) Find \(xyz\) when \(x = -2\), \(y = 5\) and \(z = 3\).
(iii) If \(p = 9\), \(q = -3\) and \(r = 4\), evaluate \(\dfrac{p - q}{r}\).
Solution
\(3 \times 4 - 2 \times (-5)\)\(=\)\(12 + 10 = 22\)
\((-2) \times 5 \times 3\)\(=\)\(-30\)
\(\dfrac{9 - (-3)}{4}\)\(=\)\(\dfrac{12}{4} = 3\)
3×4−2×(−5)=12+10=22,(−2)×5×3=−30,(9−(−3))/4=12/4=3

(i) \(22\) (subtracting a negative is adding)
(ii) \(-30\)
(iii) \(3\) (work out the top first)

Example 4 — Squares of negatives
(i) Find \(3b^2\) when \(b = -2\).
(ii) Find \(z^2 - 4z + 1\) when \(z = -5\).
Solution

Square first, with the negative in brackets: \((-2)^2 = 4\) and \((-5)^2 = 25\).

\(3 \times (-2)^2\)\(=\)\(12\)
\((-5)^2 - 4 \times (-5) + 1\)\(=\)\(46\)
3×(−2)=3×4=12,(−5)−4×(−5)+1=25+20+1=46

(i) \(3 \times 4 = 12\). Only \(b\) is squared, not \(3b\).
(ii) \(25 + 20 + 1 = 46\)

Example 5 — Formulas
(i) A school excursion costs \(C = 150 + 12n\) dollars for \(n\) students. Find the cost for \(28\) students.
(ii) The area of a rectangle is \(A = lb\). Find the area when \(l = 8\) m and \(b = 7\) m.
(iii) Use \(F = \dfrac{9}{5}C + 32\) to find \(F\) when \(C = 25\).
Solution
\(150 + 12 \times 28\)\(=\)\(150 + 336 = 486\)
\(8 \times 7\)\(=\)\(56\)
\(\dfrac{9}{5} \times 25 + 32\)\(=\)\(45 + 32 = 77\)
150+12×28=486,8×7=56,9/5×25+32=45+32=77

(i) \(\$486\)
(ii) \(56\) m²
(iii) \(F = 77\)

Example 6 — A circle
The circumference of a circle is \(C = 2\pi r\). Use \(\pi \approx \dfrac{22}{7}\) to find the circumference of a circle with radius \(14\) cm.
Solution
\(C\)\(=\)\(2 \times \dfrac{22}{7} \times 14\)
\(\)\(=\)\(2 \times 22 \times 2\)
\(\)\(=\)\(88\)
C =2×22/7×14=2×22×2=88

Divide first: \(14 \div 7 = 2\). The circumference is about \(88\) cm.

Example 7 — Checking an expansion
Lily says \(5(x + 3) = 5x + 3\). Substitute \(x = 2\) into both sides to test her answer.
Solution
\(5 \times (2 + 3)\)\(=\)\(25\)
\(5 \times 2 + 3\)\(=\)\(13\)
\(5 \times 2 + 15\)\(=\)\(25\)
5×(2+3)=25,5×2+3=13,5×2+15=25

\(25 \ne 13\), so Lily is wrong: she did not multiply the \(3\) by \(5\). The correct expansion \(5x + 15\) also gives \(25\).

Common pitfalls

Reading 3x as a two-digit number. When \(x = 5\), \(3x = 3 \times 5 = 15\), not \(35\).
Leaving out the brackets for a negative. When \(a = -3\), \(a^2 = (-3)^2 = 9\), not \(-9\).
Squaring too much. When \(b = -2\), \(3b^2 = 3 \times 4 = 12\), not \((3b)^2 = 36\).
Dividing before the top is finished. \(\dfrac{9 - (-3)}{4}\): work out \(9 + 3 = 12\) first, then divide.

Frequently asked questions

What does substitution mean in algebra?

Replacing each pronumeral with a number and working out the value.

What is the value of 3x when x = 5?

\(3 \times 5 = 15\).

Why do negative numbers go in brackets?

So the sign stays with the number: \((-3)^2 = 9\).

What is a formula?

A rule that links pronumerals, such as \(P = 2l + 2w\).

How can substitution check an expansion?

Substitute the same number into both forms. If the values differ, the expansion is wrong.