Expanding brackets and substitution – Year 7 maths lesson to download as PDF

Example 1

Expand \(4(x+5)\).

\(4\times x+4\times5=4x+20\).

Example 2

Expand \(3x(2x-7)\).

\(3x\times2x=6x^2\) and \(3x\times7=21x\). The answer is \(6x^2-21x\).

Example 3

Expand \(-5(2a-4)\).

\(-5\times2a=-10a\) and \(-5\times(-4)=+20\). The answer is \(-10a+20\).

Example 4

Expand and simplify \(3(x+4)+2(x-1)\).

\(3x+12+2x-2=5x+10\).

Example 5

Expand and simplify \(5(2y+1)-3(y+4)\).

\(10y+5-3y-12=7y-7\).

Example 6

Use \(v=u+at\) to find \(v\) when \(u=5\), \(a=2\) and \(t=3\).

\(v=5+2\times3=5+6=11\).

Example 7

Use \(C=\dfrac{5}{9}(F-32)\) to convert \(F=68\) to Celsius.

\(C=\dfrac{5}{9}\times(68-32)=\dfrac{5}{9}\times36=20^{\circ}\)C.

Example 8

A number is \(n\). Write expressions for: 5 more than the number, 3 times the number, and twice the number plus 7, all multiplied by 4.

\(n+5\), \(3n\) and \(4(2n+7)\), which expands to \(8n+28\).

Example 9

A rectangle has width \(w\) and length \(w+4\). Write and expand an expression for its area.

Area \(=w(w+4)=w^2+4w\).

Example 10

Show that \(2(x+5)\) and \(2x+10\) are equal when \(x=4\).

\(2(4+5)=2\times9=18\) and \(2\times4+10=18\). They are equal.

Example 11

Expand and simplify \(2(3x+1)-4(x-2)\).

\(6x+2-4x+8\). Careful with the sign: \(-4\times(-2)=+8\). Collect like terms: \(2x+10\).

Example 12

Use \(s=ut+\dfrac{1}{2}at^2\) to find \(s\) when \(u=3\), \(t=4\) and \(a=2\).

\(s=3\times4+\dfrac{1}{2}\times2\times4^2=12+16=28\). Indices come before multiplication.

Worked example in context

The cost of hiring a canoe is £6 plus £2.50 per hour. Write a formula and find the cost for 4 hours.

The formula is \(C=6+2.5h\). For \(h=4\): \(C=6+10=16\). The cost is £16.

Example 1

Expand \(4x(3x-2y+5)\).

The term outside the bracket is \(4x\) and there are three terms inside, so we need three multiplications. First, \(4x\times3x=12x^{2}\), because \(4\times3=12\) and \(x\times x=x^{2}\). Second, \(4x\times(-2y)=-8xy\); a positive times a negative gives a negative. Third, \(4x\times5=20x\). Putting the three results together gives \(12x^{2}-8xy+20x\). None of these are like terms, so the expression cannot be simplified any further.

Example 2

Expand and simplify \(5(2x+3)-2(4x-1)\).

We expand each bracket separately and keep track of the signs. The first bracket gives \(10x+15\). For the second bracket the number outside is \(-2\), so \(-2\times4x=-8x\) and \(-2\times(-1)=+2\). Now we write everything in one line: \(10x+15-8x+2\). Collecting the x terms gives \(10x-8x=2x\), and collecting the numbers gives \(15+2=17\). The answer is \(2x+17\). Checking with \(x=1\): the original is \(5\times5-2\times3=25-6=19\), and \(2+17=19\), so the answer is correct.

Example 3

Use the formula \(A=\dfrac{1}{2}(a+b)h\) to find the area of a trapezium with parallel sides 6 cm and 10 cm and height 4 cm.

We replace each letter by its value: \(a=6\), \(b=10\) and \(h=4\). Brackets come first in the order of operations, so \(a+b=16\). Then we multiply: \(\dfrac{1}{2}\times16\times4=32\). The area is 32 cm². Writing the substituted formula before calculating, \(A=\dfrac{1}{2}(6+10)\times4\), makes your working easy to follow and earns method marks even if you slip later.

Example 4

Factorise fully \(12x^{2}-18x\).

We look at the numbers and the letters separately. The highest common factor of 12 and 18 is 6. Both terms also contain at least one x, so x is a common factor too. Together the highest common factor is \(6x\). Dividing each term by \(6x\) gives \(12x^{2}\div6x=2x\) and \(-18x\div6x=-3\). So \(12x^{2}-18x=6x(2x-3)\). If you had only taken out 6, giving \(6(2x^{2}-3x)\), the expression would be factorised but not fully, because \(2x^{2}-3x\) still has the common factor x.