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# Chapter 5: Simple equations

## 5.1 Revise simple equation solving

Last year you learnt that an algebraic equation is a mathematical sentence that contains a combination of numbers, symbols, variables and mathematical operators. An algebraic equation always has an equals sign. You also learnt that we solve an equation when we find the value for the variable that makes the equation true. This is called the solution of the equation. For example, the number 2 is the solution of the equation $3+x=5$.

algebraic equation An algebraic equation is a mathematical sentence that contains a combination of numbers, symbols, variables and mathematical operators. It always has an equals sign.

solve When we solve an equation, we find the value of the variable that makes the equation true.

solution The solution of an equation is the value of the variable that makes the equation true.

The following worked example shows how to solve a simple equation.

### Worked example 5.1: Solving a simple equation and checking the solution

Solve the following equation and check the solution:

1. Step 1: Expand the brackets using the distributive law. If there are like terms on either side of the equation, collect them.

\begin{align} 2(x+4)+x&=-2(x+1)+20 \\ 2x+8+x&=-2x-2+20 \\ 3x+8&=-2x+18 \end{align}
2. Step 2: Use additive inverses to arrange the equation so that all the terms with variables are on the left-hand side and all the constant terms are on the right-hand side.

Remember to do exactly the same on both sides of the equation.

\begin{align} 2(x+4)+x&=-2(x+1)+20 \\ 2x+8+x&=-2x-2+20 \\ 3x+8&=-2x+18 \\ 3x+8{\color{red}{\;-\;8}}{\color{green}{\;+\;2x}} &=-2x{\color{green}{\;+\;2x}}+18 {\color{red}{\;-\;8}} \\ \end{align}
3. Step 3: Simplify both sides of the equation.

\begin{align} 2(x+4)+x&=-2(x+1)+20 \\ 2x+8+x&=-2x-2+20 \\ 3x+8&=-2x+18 \\ 3x+8{\color{red}{\;-\;8}}{\color{green}{\;+\;2x}} &=-2x{\color{green}{\;+\;2x}}+18 {\color{red}{\;-\;8}} \\ 3x+2x &=18-8 \\ 5x &=10 \end{align}
4. Step 4: If the variable has a coefficient, apply the multiplicative inverse of the coefficient. Remember to do this on both sides of the equation.

\begin{align} 2(x+4)+x&=-2(x+1)+20 \\ 2x+8+x&=-2x-2+20 \\ 3x+8&=-2x+18 \\ 3x+8{\color{red}{\;-\;8}}{\color{green}{\;+\;2x}} &=-2x{\color{green}{\;+\;2x}}+18 {\color{red}{\;-\;8}} \\ 3x+2x &=18-8 \\ 5x &=10 \\ {\color{orange}{\frac{1}{5}\times}} \frac{5x}{1}&={\color{orange}{\frac{1}{5}\times}} \frac{10}{1} \\ \end{align}
5. Step 5: Simplify both sides of the equation.

\begin{align} 2(x+4)+x&=-2(x+1)+20 \\ 2x+8+x&=-2x-2+20 \\ 3x+8&=-2x+18 \\ 3x+8{\color{red}{\;-\;8}}{\color{green}{\;+\;2x}} &=-2x{\color{green}{\;+\;2x}}+18 {\color{red}{\;-\;8}} \\ 3x+2x &=18-8 \\ 5x &=10 \\ {\color{orange}{\frac{1}{5}\times}} \frac{5x}{1}&={\color{orange}{\frac{1}{5}\times}} \frac{10}{1} \\ \frac{5x}{5}&=\frac{10}{5} \\ \therefore x&=2 \end{align}
6. Step 6: Check your solution. Replace the variable on the left-hand side (LHS) of the original equation with the solution. Work out the value of the left-hand side.

\begin{align} \text{LHS}&=2(x+4)+x \\ &=2(2+4)+2 \\ &=2(6)+2 \\ &=12+2\\ &=14 \end{align}
7. Step 7: Repeat Step 6 for the right-hand side (RHS) of the original equation.

\begin{align} \text{RHS}&=-2(x+1)+20 \\ &=-2(2+1)+20 \\ &=-2(3)+20 \\ &=-6+20\\ &=14 \end{align}
8. Step 8: Compare your answers from Steps 6 and 7. If they are the same, your solution is correct. If not, you made a mistake somewhere. Solve the equation again.

$\text{LHS}=\text{RHS}$
The solution is correct.

### Exercise 5.1: Solve simple equations and check the solutions

Solve each of the following equations and check the solutions.

1. \begin{align} 13+3&=2x-3+3 \newline 16&=2x \newline \frac{16}{2}&=\frac{2x}{2} \newline 8&=x \newline \therefore x&=8 \end{align}

$8=x$ and $x=8$ means exactly the same.

Checking:

\begin{align} 13&=2(8)-3 \newline 13&=16-3 \newline 13&=13 \newline \therefore \text{LHS}&=\text{RHS} \end{align}
2. \begin{align} 39&=21+7q-4q \newline 39&=21+3q \newline 39-21&=21-21+3q \newline 18&=3q \newline \frac{18}{3}&=\frac{3q}{3} \newline 6&=q \newline \therefore q&=6 \end{align}

Checking:

\begin{align} 39&=21+5(6)-4(6)+2(6) \newline 39&=21+30-24+12 \newline 39&=63-24 \newline 39&=39 \newline \therefore \text{LHS}&=\text{RHS} \end{align}
3. \begin{align} 2(y+3)&=10 \\ 2y+6&=10 \\ 2y+6-6&=10-6 \\ 2y&=4 \\ \frac{2y}{2}&=\frac{4}{2} \\ \therefore y&=2 \end{align}

Checking:

\begin{align} 2(2+3)&=10 \\ 2(5)&=10 \\ 10&=10 \\ \therefore \text{LHS}&=\text{RHS} \end{align}
4. \begin{align} 2(y+1)&=6(y-1) \\ 2y+2&=6y-6 \\ 2y+2-2-6y &= 6y-6y-6-2 \\ -4y&=-8\\ \frac{-4y}{-4}&=\frac{-8}{-4} \\ \therefore y&=2 \end{align}

Checking:

\begin{align} 2(y+1)&=6(y-1) \\ 2(2+1)&=6(2-1) \\ 2(3)&=6(1) \\ 6&=6 \\ \therefore \text{LHS}&=\text{RHS} \end{align}
5. \begin{align} 2(5a-3)-5(7a-3)&=2a \\ 10a-6-35a+15&=2a \\ -25a+9 &= 2a \\ -25a+9-9-2a&=2a-2a-9\\ -27a &= -9 \\ \frac{-27a}{-27}&=\frac{-9}{-27} \\ \therefore a&=\frac{1}{3} \end{align}

Checking:

\begin{align} 2(5a-3)-5(7a-3)&=2a \\ 2(\frac{5}{1} \times \frac{1}{3}-3) - 5(\frac{7}{1} \times\frac{1}{3}-3)&=\frac{2}{1}\times\frac{1}{3} \\ 2(\frac{5}{3}-\frac{9}{3})-5(\frac{7}{3}-\frac{9}{3})&=\frac{2}{3}\\ \frac{2}{1}(\frac{-4}{3})-\frac{5}{1}(\frac{-2}{3})&=\frac{2}{3}\\ \frac{-8}{3}+\frac{10}{3}&=\frac{2}{3}\\ \frac{2}{3}&=\frac{2}{3} \\ \therefore \text{LHS}&=\text{RHS} \end{align}

## 5.2 Solving equations with fractions

Last year you learnt how to solve simple equations with fractions. The following worked example shows how to solve an equation with fractions that have constants as denominators.

### Worked example 5.2: Solving an equation with fractions that have constants as denominators

Solve the following equation and check the solution:

1. Step 1: Find the LCD of the denominators.

The LCD of 3 and 6 is 6.

2. Step 2: Multiply all the terms of the equation with the LCD from Step 1.

\begin{align} {\color{red}{\frac{6}{1}\times}} \frac{2x-1}{3}&=\frac{5x-1}{6}{\color{red}{\times\frac{6}{1}}}\\ \end{align}

Note that we are doing exactly the same on both sides of the equation.

3. Step 3: For each term of the equation, find the HCF of the denominator of the term and the LCD from Step 1. Divide by the HCF.

The HCF of 6 and 3 is 3. Therefore, $\frac{6}{3}=2$, $\frac{3}{3}=1$ and $\frac{6}{6}=1$.

\require{cancel} \begin{align} {\color{red}{\frac{6}{1}\times}} \frac{2x-1}{3}&=\frac{5x-1}{6}{\color{red}{\times\frac{6}{1}}}\\ \frac{\color{green}{\cancel{6}^{2}}}{1} \times \frac{2x-1}{\color{green}{\cancel{3}^1}} &=\frac{5x-1}{\color{green}{\cancel{6}^1}} \times \frac{\color{green}{\cancel{6}^1}}{1}\\ \end{align}
4. Step 4: Multiply the factors that remain in each term.

\begin{align} {\color{red}{\frac{6}{1}\times}} \frac{2x-1}{3}&=\frac{5x-1}{6}{\color{red}{\times\frac{6}{1}}}\\ \frac{\color{green}{\cancel{6}^{2}}}{1} \times \frac{2x-1}{\color{green}{\cancel{3}^1}} &=\frac{5x-1}{\color{green}{\cancel{6}^1}} \times \frac{\color{green}{\cancel{6}^1}}{1}\\ 2(2x-1)&=5x-1 \end{align}

Remember that a number or variable remains the same if you divide or multiply it by 1, for example $2x\times 1=2x$ and $\frac{5x}{1}=5x$.

5. Step 5: Solve the equation.

\begin{align} 2(2x-1)&=5x-1 \\ 4x-2&=5x-1 \\ 4x-2+2-5x&=5x-5x-1+2 \\ -x&=1 \\ \frac{-x}{-1}&=\frac{1}{-1} \\ \therefore x&=-1 \end{align}
6. Step 6: Check your solution. Remember to substitute the solution for the variable in the original equation.

\begin{align} \frac{2x-1}{3}&=\frac{5x-1}{6} \\ \frac{2(-1)-1}{3}&=\frac{5(-1)-1}{6} \\ \frac{-2-1}{3}&=\frac{-5-1}{6}\\ \frac{-3}{3}&=\frac{-6}{6} \\ -1&=-1 \\ \therefore \text{LHS}&=\text{RHS} \end{align}

### Exercise 5.2: Solve equations with fractions that have constants as denominators

Solve each of the following equations and check the solutions.

1. \begin{align} \frac{2a}{35}&=\frac{4}{5} \\ \frac{35}{1}\times \frac{2a}{35}&=\frac{4}{5}\times \frac{35}{1} \\ \frac{\cancel{35}^1}{1}\times \frac{2a}{\cancel {35}^1}&=\frac{4}{\cancel 5^1}\times \frac{\cancel{35}^7}{1} \\ 2a&=28\\ \frac{2a}{2}&=\frac{28}{2} \\ \therefore a&=14 \end{align}

Checking:

\begin{align} \frac{2(14)}{35}&=\frac{4}{5} \\ \frac{28}{35}&=\frac{4}{5} \\ \frac{28\div 7}{35 \div 7}&=\frac{4}{5} \\ \frac{4}{5}&=\frac{4}{5} \\ \therefore \text{LHS}&=\text{RHS} \end{align}
2. \begin{align} 1+\frac{p}{4}&=\frac{p}{5} \\ \frac{20}{1}\times 1+\frac{20}{1}\times \frac{p}{4}&=\frac{p}{5}\times \frac{20}{1} \\ \frac{20}{1}+\frac{\cancel{20}^5}{1}\times \frac{p}{\cancel 4^1}&=\frac{p}{\cancel 5^1}\times \frac{\cancel{20}^4}{1} \\ 20+5p&=4p\\ 20-20+5p-4p&=4p-4p-20 \\ \therefore p&=-20 \end{align}

Checking:

\begin{align} 1+\frac{-20}{4}&=\frac{-20}{5} \\ 1-5&=-4 \\ -4&=-4 \\ \therefore \text{LHS}&=\text{RHS} \end{align}
3. \begin{align} \frac{2n-1}{4}&=0 \\ \frac{4}{1} \times \frac{2n-1}{4}&=0 \times \frac{4}{1} \\ \frac{\cancel 4^1}{1} \times \frac{2n-1}{\cancel 4^1}&=0 \times \frac{4}{1} \\ 2n-1&=0\\ 2n-1+1&=0+1 \\ 2n&=1 \\ \frac{2n}{2}&=\frac{1}{2} \\ \therefore n&=\frac{1}{2} \end{align}

Checking:

\begin{align} \frac{\frac{2}{1}(\frac{1}{2})-1}{4}&=0 \\ \frac{1-1}{4}&=0 \\ \frac{0}{4}&=0\\ 0&=0 \\ \therefore \text{LHS}&=\text{RHS} \end{align}
4. \begin{align} \frac{x-1}{15}&=\frac{1}{3} \\ \frac{15}{1}\times\frac{x-1}{15}&=\frac{1}{3}\times\frac{15}{1} \\ \frac{\cancel{15}^1}{1}\times\frac{x-1}{\cancel{15}^1}&=\frac{1}{\cancel 3^1}\times\frac{\cancel{15}^5}{1} \\ x-1&=5\\ x-1+1&=5+1 \\ \therefore x&=6 \end{align}

Checking:

\begin{align} \frac{6-1}{15}&=\frac{1}{3} \\ \frac{5\div 5}{15\div 5}&=\frac{1}{3} \\ \frac{1}{3}&=\frac{1}{3}\\ \therefore \text{LHS}&=\text{RHS} \end{align}
5. \begin{align} \frac{b-1}{2}&=\frac{b+1}{4} \\ \frac{4}{1}\times\frac{b-1}{2}&=\frac{b+1}{4}\times\frac{4}{1} \\ \frac{\cancel 4^2}{1}\times\frac{b-1}{\cancel 2^1}&=\frac{b+1}{\cancel 4^1}\times\frac{\cancel 4^1}{1} \\ 2(b-1)&=b+1\\ 2b-2&=b+1 \\ 2b-b-2+2&=b-b+1+2\\ \therefore b&=3 \\ \end{align}

Checking:

\begin{align} \frac{3-1}{2}&=\frac{3+1}{4} \\ \frac{2}{2}&=\frac{4}{4} \\ 1&=1 \\ \therefore \text{LHS}&=\text{RHS} \end{align}

The following worked example shows how to solve an equation with fractions that have variables in their denominators.

### Worked example 5.3: Solving an equation with fractions that have variable denominators

Solve the following equation and check the solution:

1. Step 1: Find the LCD of the denominators.

Find the LCD of the numbers in the denominators:

• The prime factors of the denominators are: $3; \; 2^2\times3; \; 2^2$
• The LCD is: $3\times 2^2=12$

Find the LCD of the variables in the denominators:

• The only variable is: $x$
2. Step 2: Multiply all the terms of the equation by the LCD from Step 1.

\begin{align} {\color{red}{\frac{12x}{1}\times}} \frac{2x+1}{3}+{\color{red}{\frac{12x}{1}\times}} \frac{5}{12x}&=\frac{3}{4x}{\color{red}{\times\frac{12x}{1}}} +\frac{2x}{3}{\color{red}{\times\frac{12x}{1}}}\\ \end{align}

Note that we are multiplying each term, on both sides of the equation, by the LCD.

3. Step 3: For each term of the equation, find the HCF of the denominator of the term and the LCD from Step 1. Divide by the HCF.

\begin{align} {\color{red}{\frac{\cancel{12x}^{4x}}{1}\times}} \frac{2x+1}{\cancel 3^1}+{\color{red}{\frac{\cancel{12x}^1}{1}\times}} \frac{5}{\cancel{12x}^1}&=\frac{3}{\cancel{4x}^1}{\color{red}{\times\frac{\cancel{12x}^3}{1}}} + \frac{2x}{\cancel 3^1}{\color{red}{\times\frac{\cancel{12x}^{4x}}{1}}}\\ \end{align}
4. Step 4: Multiply the factors that remain in each term.

5. Step 5: Solve the equation.

\begin{align} 4x(2x+1)+5&=9+8x^2 \\ 8x^2+4x+5&=9+8x^2 \\ 8x^2-8x^2+4x+5-5&=9+8x^2-8x^2-5 \\ 4x&=4 \\ \frac{4x}{4}&=\frac{4}{4} \\ \therefore x&=1 \end{align}
6. Step 6: Check your solution. Remember to replace the variable with the solution in the original equation.

\begin{align} \frac{2(1)+1}{3}+\frac{5}{12(1)}&=\frac{3}{4(1)}+\frac{2(1)}{3} \\ \frac{3}{3}+\frac{5}{12}&=\frac{3}{4}+\frac{2}{3} \\ \frac{12}{12}+\frac{5}{12}&=\frac{9}{12}+\frac{8}{12}\\ \frac{17}{12}&=\frac{17}{12} \\ \therefore \text{LHS}&=\text{RHS} \end{align}

### Exercise 5.3: Solve equations with fractions that have variable denominators

Solve each of the following equations and check the solutions.

1. \begin{align} \frac{6a}{1}\times\frac{8}{6a}&=\frac{2}{3} \times \frac{6a}{1} \\ \frac{\cancel{6a}^1}{1}\times\frac{8}{\cancel {6a}^1}&=\frac{2}{\cancel 3^1} \times \frac{\cancel{6a}^{2a}}{1}\\ 8&=4a \\ \frac{8}{4}&=\frac{4a}{4}& \\ 2&=a \\ \therefore a&=2 \end{align}

Checking:

\begin{align} \frac{8}{6(2)}&=\frac{2}{3} \\ \frac{8}{12}&=\frac{2}{3} \\ \frac{2}{3}&=\frac{2}{3} \\ \therefore \text{LHS}&=\text{RHS} \end{align}
2. \begin{align} \frac{2y}{1}\times\frac{3}{2}-\frac{2y}{1}\times\frac{4}{y}&=1\times\frac{2y}{1} \\ \frac{\cancel{2y}^y}{1}\times\frac{3}{\cancel 2^1}-\frac{\cancel{2y}^2}{1}\times\frac{4}{\cancel y^1}&=\frac{2y}{1} \\ 3y-8 &= 2y \\ 3y-2y-8+8&=2y-2y+8\\ \therefore y&=8 \\ \end{align}

Checking:

\begin{align} \frac{3}{2}-\frac{4}{8}&=1 \\ \frac{3}{2}-\frac{1}{2}&=1 \\ \frac{2}{2}&=1 \\ 1&=1 \\ \therefore \text{LHS}&=\text{RHS} \end{align}
3. \begin{align} \frac{10x}{1}\times\frac{3}{2x}+\frac{10x}{1}\times\frac{1}{5}&=\frac{1}{x}\times\frac{10x}{1} \\ \frac{\cancel{10x}^5}{1}\times\frac{3}{\cancel{2x}^1}+\frac{\cancel{10x}^{2x}}{1}\times\frac{1}{\cancel 5^1}&=\frac{1}{\cancel x^1}\times\frac{\cancel{10x}^{10}}{1} \\ 15+2x &= 10 \\ 15-15+2x &= 10-15\\ 2x&=-5 \\ \frac{2x}{2}&=\frac{-5}{2} \\ \therefore x&=-\frac{5}{2} \end{align}

Checking:

\begin{align} \frac{3}{\frac{2}{1}\times \frac{-5}{2}}+\frac{1}{5}&=\frac{1}{\frac{-5}{2}} \\ \frac{3}{\frac{\cancel 2^1}{1}\times \frac{-5}{\cancel 2^1}}+\frac{1}{5}&=\frac{1}{\frac{-5}{2}}\\ \frac{3}{1}\times \frac{1}{-5}+\frac{1}{5}&=\frac{1}{1}\times\frac{2}{-5} \\ -\frac{3}{5}+\frac{1}{5}&=-\frac{2}{5} \\ -\frac{2}{5}&=-\frac{2}{5} \\ \therefore \text{LHS}&=\text{RHS} \end{align}

Remember that dividing by a fraction is the same as multiplying by the reciprocal of the fraction. For example $\frac{1}{\frac{3}{5}}=1\times\frac{5}{3}$.

4. \begin{align} \frac{15b}{1}\times\frac{7}{15}+\frac{15b}{1}\times\frac{1}{b}&=\frac{1}{15}\times\frac{15b}{1}-\frac{1}{5b}\times\frac{15b}{1} \\ \frac{\cancel{15b}^b}{1}\times\frac{7}{\cancel{15}^1}+\frac{\cancel{15b}^{15}}{1}\times\frac{1}{\cancel b^1}&=\frac{1}{\cancel{15}^1}\times\frac{\cancel{15b}^b}{1}-\frac{1}{\cancel{5b}^1}\times\frac{\cancel{15b}^3}{1} \\ 7b+15&=b-3 \\ 7b+15-15-b&=b-3-b-15\\ 6b&=-18 \\ \frac{6b}{6}&=\frac{-18}{6} \\ \therefore b&=-3 \end{align}

Checking:

\begin{align} \frac{7}{15}+\frac{1}{-3}&=\frac{1}{15}-\frac{1}{5(-3)} \\ \frac{7}{15}-\frac{5}{15}&=\frac{1}{15}+\frac{1}{15}\\ \frac{2}{15}&=\frac{2}{15} \\ \therefore \text{LHS}&=\text{RHS} \end{align}

Remember that subtracting a number is the same as adding the additive inverse of the number. For example $\frac{1}{15}-(-\frac{1}{15})=\frac{1}{15}+\frac{1}{15}$.

5. \begin{align} \frac{72x}{1}\times\frac{1}{4x}+\frac{72x}{1}\times\frac{5}{9}-\frac{72x}{1}\times\frac{3}{6x}&=\frac{5}{6x}\times\frac{72x}{1}+\frac{1}{72}\times\frac{72x}{1} \\ \frac{\cancel{72x}^{18}}{1}\times\frac{1}{\cancel{4x}^1}+\frac{\cancel{72x}^{8x}}{1}\times\frac{5}{\cancel 9^1}-\frac{\cancel{72x}^{12}}{1}\times\frac{3}{\cancel{6x}^1}&=\frac{5}{\cancel{6x}^1}\times\frac{\cancel{72x}^{12}}{1}+\frac{1}{\cancel{72}^1}\times\frac{\cancel{72x}^x}{1} \\ 18+40x-36&=60+x \\ 18-18+40x-36+36-x&=60+x-x-18+36\\ 39x&=78 \\ \frac{39x}{39}&=\frac{78}{39} \\ \therefore x&=2 \end{align}

Checking:

\begin{align} \frac{1}{4(2)}+\frac{5}{9}-\frac{3}{6(2)}&=\frac{5}{6(2)}+\frac{1}{72} \\ \frac{1}{8}+\frac{5}{9}-\frac{3}{12}&=\frac{5}{12}+\frac{1}{72}\\ \frac{9}{72}+\frac{40}{72}-\frac{18}{72}&=\frac{30}{72}+\frac{1}{72}\\ \frac{31}{72}&=\frac{31}{72} \\ \therefore \text{LHS}&=\text{RHS} \end{align}

In Chapter 4 you learnt that a polynomial is an algebraic expression with two or more terms, of which one can be a constant, and the other terms consist of coefficients with variables raised to positive powers. A binomial has two terms, for example $2x+3$. The following worked example shows how to solve an equation with fractions that have binomial denominators.

### Worked example 5.4: Solving an equation with fractions that have binomial denominators

Solve the following equation and check the solution:

1. Step 1: Find the LCD of the denominators.

Regard each denominator as a factor of the LCD. Therefore, the LCD of the two denominators is: $(x-3)(2x+1)$.

2. Step 2: Multiply all the terms of the equation with the LCD from Step 1.

\begin{align} {\color{red}{\frac{(x-3)(2x+1)}{1}\times}} \frac{1}{x-3}&=\frac{3}{2x+1}{\color{red}{\times\frac{(x-3)(2x+1)}{1}}}\\ \end{align}
3. Step 3: For each term of the equation, find the HCF of the denominator of the term and the LCD from Step 1. Divide by the HCF.

\begin{align} {\color{red}{\frac{\cancel{(x-3)}^1(2x+1)}{1}\times}} \frac{1}{\cancel{x-3}^1}&=\frac{3}{\cancel{2x+1}^1}{\color{red}{\times\frac{(x-3)\cancel{(2x+1)}^1}{1}}}\\ \end{align}
4. Step 4: Multiply the factors that remain in each term.

5. Step 5: Solve the equation.

\begin{align} 2x+1&=3(x-3) \\ 2x+1&=3x-9 \\ 2x-3x+1-1&=3x-3x-9-1 \\ -x&=-10 \\ \frac{-x}{-1}&=\frac{-10}{-1} \\ \therefore x&=10 \end{align}
6. Step 6: Check your solution. Remember to substitute the solution for the variable in the original equation.

\begin{align} \frac{1}{10-3}&=\frac{3}{2(10)+1} \\ \frac{1}{7}&=\frac{3}{21} \\ \frac{1}{7}&=\frac{1}{7}\\ \therefore \text{LHS}&=\text{RHS} \end{align}

### Exercise 5.4: Solve equations with fractions that have binomial denominators

Solve each of the following equations and check the solutions.

1. \begin{align} \frac{2(a-1)}{1}\times\frac{2a+1}{a-1}&=\frac{1}{2}\times\frac{2(a-1)}{1} \\ \frac{2\cancel{(a-1)}^1}{1}\times\frac{2a+1}{\cancel{a-1}^1}&=\frac{1}{\cancel 2^1}\times\frac{\cancel 2^1(a-1)}{1} \\ 2(2a+1)&=1(a-1) \\ 4a+2&=a-1\\ 4a+2-2-a&=a-a-1-2 \\ 3a&=-3 \\ \frac{3a}{3}&=\frac{-3}{3}\\ \therefore a&=-1 \end{align}

Checking:

\begin{align} \frac{2(-1)+1}{-1-1}&=\frac{1}{2} \\ \frac{-1}{-2}&=\frac{1}{2}\\ \frac{1}{2}&=\frac{1}{2}\\ \therefore \text{LHS}&=\text{RHS} \end{align}
2. \begin{align} \frac{(3b+2)(3b-1)}{1}\times\frac{2}{3b+2}&=\frac{1}{3b-1}\times\frac{(3b+2)(3b-1)}{1} \\ \frac{\cancel{(3b+2)}^1(3b-1)}{1}\times\frac{2}{\cancel{3b+2}^1}&=\frac{1}{\cancel{3b-1}^1}\times\frac{(3b+2)\cancel{(3b-1)}^1}{1} \\ 2(3b-1)&=1(3b+2) \\ 6b-2&=3b+2\\ 6b-2+2-3b&=3b-3b+2+2 \\ 3b&=4 \\ \frac{3b}{3}&=\frac{4}{3}\\ \therefore b&=\frac{4}{3} \end{align}

Checking:

\begin{align} \frac{2}{\frac{3}{1}\times\frac{4}{3}+2}&=\frac{1}{\frac{3}{1}\times\frac{4}{3}-1} \\ \frac{2}{4+2}&=\frac{1}{4-1}\\ \frac{2}{6}&=\frac{1}{3}\\ \frac{1}{3}&=\frac{1}{3}\\ \therefore \text{LHS}&=\text{RHS} \end{align}
3. \begin{align} \frac{(y^2+1)(2y-1)}{1}\times\frac{2y}{y^2+1}&=\frac{4}{2y-1}\times\frac{(y^2+1)(2y-1)}{1} \\ \frac{\cancel{(y^2+1)}^1(2y-1)}{1}\times\frac{2y}{\cancel{y^2+1}^1}&=\frac{4}{\cancel{2y-1}^1}\times\frac{(y^2+1)\cancel{(2y-1)}^1}{1} \\ 2y(2y-1)&=4(y^2+1) \\ 4y^2-2y&=4y^2+4\\ 4y^2-4y^2-2y&=4y^2-4y^2+4 \\ -2y&=4 \\ \frac{-2y}{-2}&=\frac{4}{-2}\\ \therefore y&=-2 \end{align}

Checking:

\begin{align} \frac{2(-2)}{(-2)^2+1}&=\frac{4}{2(-2)-1} \\ \frac{-4}{4+1}&=\frac{4}{-4-1}\\ \frac{-4}{5}&=\frac{4}{-5}\\ \therefore \text{LHS}&=\text{RHS} \end{align}

Remember: $(-2)^2=-2\times -2 =+4$

4. \begin{align} \frac{y+3}{2y+1}&=\frac{3}{2} \\ \frac{2(2y+1)}{1}\times \frac{y+3}{2y+1}&=\frac{3}{2}\times\frac{2(2y+1)}{1} \\ \frac{2\cancel{(2y+1)}^1}{1}\times \frac{y+3}{\cancel{2y+1}^1}&=\frac{3}{\cancel 2^1}\times\frac{\cancel 2^1(2y+1)}{1} \\ 2(y+3)&=3(2y+1) \\ 2y+6&=6y+3\\ 2y+6-6-6y&=6y-6y+3-6 \\ -4y&=-3 \\ \frac{-4y}{-4}&=\frac{-3}{-4}\\ \therefore y&=\frac{3}{4} \end{align}

Checking:

\begin{align} \frac{\frac{3}{4}+\frac{3}{1}}{\frac{2}{1}\times\frac{3}{4}+\frac{1}{1}}&=1\frac{1}{2} \\ \frac{\frac{3}{4}+\frac{12}{4}}{\frac{6}{4}+\frac{4}{4}}&=\frac{3}{2}\\ \frac{\frac{15}{4}}{\frac{10}{4}}&=\frac{3}{2}\\ \frac{15}{4}\times\frac{4}{10}&=\frac{3}{2}\\ \frac{15}{10}&=\frac{3}{2}\\ \frac{3}{2}&=\frac{3}{2}\\ \therefore \text{LHS}&=\text{RHS} \end{align}
5. \begin{align} \frac{3}{x-8}-\frac{2}{x+3}&=0 \\ \frac{(x-8)(x+3)}{1}\times \frac{3}{x-8}-\frac{(x-8)(x+3)}{1}\times\frac{2}{x+3}&=0\times \frac{(x-8)(x+3)}{1} \\ \frac{\cancel{(x-8)}^1(x+3)}{1}\times \frac{3}{\cancel{x-8}^1}-\frac{(x-8)\cancel{(x+3)}^1}{1}\times\frac{2}{\cancel{x+3}^1}&=0 \\ 3(x+3)-2(x-8)&=0 \\ 3x+9-2x+16&=0 \\ 3x+9-2x+16&=0 \\ x+25&=0 \\ x+25-25&=-25 \\ \therefore x&=-25\\ \end{align}

Checking:

\begin{align} \frac{3}{-25-8}-\frac{2}{-25+3}&=0 \\ \frac{3}{-33}-\frac{2}{-22}&=0\\ \frac{1}{-11}+\frac{1}{11}&=0\\ 0&=0\\ \therefore \text{LHS}&=\text{RHS} \end{align}

## 5.3 Using equations to solve problems

Equations are extremely useful when solving problems that are described in words. In previous years you learnt that these steps can be used to solve a problem that is described in words:

1. Understand the problem: Read the question carefully a few times. Identify the information you have. Make sure you know what is asked. It may help to draw a picture.
2. Make a plan: Identify the unknown(s). Make one unknown $x$. If there is more than one unknown, express all of them in terms of $x$. Write an equation that you can use to solve the problem.
3. Carry out the plan: Solve the equation.
4. Look back: Check your solution(s). Make sure that you give the answers to everything that was asked.

### Worked example 5.5: Solving a problem using an equation

A mother is 30 years older than her daughter. Five years ago, the daughter's age was one seventh of the mother's age. Calculate how old the mother and her daughter are currently.

1. Step 1: Understand the problem.

• Given: Mother is 30 years older than daughter
• Given: Five years ago, daughter's age was $\frac{1}{7}$ of mother's age
• Asked: Present ages of mother and daughter
2. Step 2: Devise a plan.

• Daughter's present age: $x$
• Mother's present age: $x+30$
• Daughter's age five years ago: $x-5$
• Mother's age five years ago: $(x+30)-5=x+25$

The equation is:

3. Step 3: Carry out the plan.

\begin{align} \frac{x-5}{1} &= \frac{1}{7}\times \frac{x+25}{1} \\ \frac{x-5}{1} &= \frac{x+25}{7} \\ \frac{7}{1}\times\frac{x-5}{1} &= \frac{x+25}{\cancel 7^1}\times\frac{\cancel 7^1}{1} \\ 7(x-5)&=x+25 \\ 7x-35&=x+25 \\ 7x-x-35+35&=x-x+25+35 \\ 6x&=60 \\ \frac{6x}{6}&=\frac{60}{6} \\ \therefore x&=10 \end{align}

Therefore:

• Daughter's present age: $10$ years
• Mother's present age: $10+30=40$ years
4. Step 4: Look back.

\begin{align} x-5&=\frac{1}{7}\times (x+25)\\ 10-5&=\frac{1}{7}\times (10+25)\\ 5&=\frac{1}{7}\times \frac{35}{1} \\ 5&=\frac{1}{\cancel 7^1}\times \frac{\cancel{35}^5}{1} \\ 5&=5 \\ \therefore \text{LHS}&=\text{RHS} \end{align}

The daughter is currently 10 years old, and the mother is currently 40 years old.

### Exercise 5.5: Use equations to solve problems

1. Your class is going on a bus trip, and you want to buy snacks and water for the trip. At a certain store, packets of groundnuts cost five times more than sachets of water. If you spend $$₦\,250$$ on water sachets and $$₦\,500$$ on packets of groundnuts, you can buy 35 water sachets and packets of groundnuts altogether. Determine the cost of a water sachet and the cost of a packet of groundnuts.

Let the cost of a water sachet be $&#8358;\,x$.
Then the cost of a packet of groundnuts is $&#8358;\,5x$.

\begin{align} \frac{250}{x}+\frac{500}{5x} &=35 \\ \frac{250}{x}+\frac{100}{x} &=35 \\ \frac{350}{x} &=35 \\ \frac{x}{1}\times\frac{350}{x} &=\frac{35}{1} \times \frac{x}{1}\\ \frac{\cancel x^1}{1}\times\frac{350}{\cancel x^1} &=\frac{35x}{1} \\ 350&=35x \\ \frac{35x}{35}&=\frac{350}{35} \\ \therefore x&=10 \end{align}

The price of a water sachet is $&#8358;\,10$.
The price of a packet of groundnut is $&#8358;\,10\times5 = &#8358;\,50$.

2. The perimeter of a square is 21 units longer than half of one of its sides. Calculate its side length.

Let the side length be $x$.
Then the perimeter is $\frac{1}{2}x+21$.

\begin{align} 4s&=P \\ 4x&=\frac{1}{2}x+21 \\ \frac{4x}{1}&=\frac{x}{2}+\frac{21}{1} \\ \frac{2}{1}\times\frac{4x}{1}&=\frac{x}{\cancel 2^1}\times\frac{\cancel 2^1}{1}+\frac{21}{1}\times\frac{2}{1} \\ 8x&=x+42 \\ 8x-x&=x-x+42 \\ \frac{7x}{7}&=\frac{42}{7} \\ \therefore x&=6 \end{align}

The side length is 6 units.

3. The breadth of a rectangle is one unit longer than one third of its length. The perimeter of the rectangle is 34 units. Determine the length and breadth of the rectangle.

Let the length be $x$.
Then the breadth is $\frac{1}{3}x+1$.

\begin{align} 2l+2b&=P \\ 2x+2(\frac{1}{3}x+1)&=34 \\ 2x+\frac{2}{1}(\frac{x}{3}+1)&=34 \\ 2x+\frac{2x}{3}+2&=34\\ 2x+\frac{2x}{3}+2-2&=34-2\\ \frac{3}{1}\times\frac{2x}{1}+\frac{\cancel 3^1}{1}\times\frac{2x}{\cancel 3^1}&=\frac{32}{1}\times\frac{3}{1}\\ 6x+2x&=96 \\ 8x&=96 \\ \frac{8x}{8}&=\frac{96}{8} \\ \therefore x&=12 \end{align}

The length is 12 units.
The breadth is $\frac{1}{3}\times\frac{12}{1}+1=\frac{12}{3}+1=$5 units.

4. If you add 15 to a number, then one third of the answer is equal to double the number. Determine the number.

Let the number be $x$.

\begin{align} \frac{1}{3}(x+15)&=2x \\ \frac{x+15}{3}&=\frac{2x}{1} \\ \frac{3}{1}\times\frac{x+15}{3}&=\frac{2x}{1}\times\frac{3}{1} \\ \frac{\cancel 3^1}{1}\times\frac{x+15}{\cancel 3^1}&=\frac{2x\times 3}{1} \\ x+15&=6x \\ x-6x+15-15 &= 6x-6x-15 \\ -5x&=-15 \\ \frac{-5x}{-5}&=\frac{-15}{-5} \\ \therefore x&=3 \\ \end{align}

The number is 3.

5. Eighteen bananas and twenty seven tamarinds must be divided amongst a group of children. One child in the group does not want any bananas. The remaining children all get the same number of bananas. Two additional children join the group when the tamarinds are divided and everybody gets the same number of tamarinds. The number of bananas per child is equal to the number of tamarinds per child. How many children were in the original group?

Let the number of children in the original group be $x$.
The 18 bananas are shared amongst $x-1$ children.
The 27 tamarinds are shared amongst $x+2$ children.

\begin{align} \frac{18}{x-1}&=\frac{27}{x+2} \\ \frac{\cancel{(x-1)}^1(x+2)}{1}\times\frac{18}{\cancel{x-1}^1}&=\frac{27}{\cancel{x+2}^1}\times \frac{(x-1)\cancel{(x+2)}^1}{1} \\ 18(x+2)&=27(x-1) \\ 18x+36 &= 27x-27 \\ 18x-27x+36-36 &= 27x-27x-27-36 \\ -9x&=-63\\ \frac{-9x}{-9}&=\frac{-63}{-9} \\ \therefore x&=7 \\ \end{align}

The original group had 7 children.

## 5.4 Summary

• An algebraic equation is a mathematical sentence that contains a combination of numbers, symbols, variables and mathematical operators. It always has an equals sign.
• The solution of an equation is the value of the variable that makes the equation true.
• When we solve an equation, we find the value of the variable that makes the equation true.
• To solve equations with fractions we can follow this method:
• Find the LCD of the denominators.
• Multiply all the terms of the equation by the LCD of the denominators.
• For each term of the equation, divide by the HCF of the denominators of that term and the LCD.
• Multiply the factors that remain in each term.
• Solve the equation.
• Check the solution.
• We can follow these steps to solve a problem that is described in words:
• Understand the problem, by reading it carefully.
• Make a plan, by identifying the unknown(s) and writing an equation.
• Carry out the plan, by solving the equation.
• Look back, by checking your solution.