7.3 Solving equations by inspection
Searching for the solution of an equation by using tables or by narrowing down to the possible solution is called solution by inspection.
When a certain number is the solution of an equation, we say that the number satisfies the equation. For example, \(x = 4\) satisfies the equation \(3x = 12\) because \(3 \times 4 = 12\).
- An equation is like a scale, but coefficients and variables are balanced instead of balancing weights or numbers.
- A variable is an unknown number represented by the letters of the alphabet. In the example above, \(x\) is a variable.
- Constants are numbers because they constantly have the same value. In the example \(3x = 12\), the constants are \(3\) and \(12\).
Equations in algebra are based on the principle of calculating the value in the place of a variable. Which value will make the equation true?
The value \(x = 4\) makes the equation \(3x = 12\) true.
Worked Example 7.4: Finding the solution by inspection
What makes the following equation true?
\[12x + 14 = 50\]Inspect both sides of the equation.
Left-hand side: \(12x + 14\)
- A variable \(x\) is multiplied by \(12\).
- A constant number \(14\) is added.
Right-hand side: \(50\)
- We know that \(50 = 36 + 14\)
Balance the equation.
We can write the equation like this:
\[12x + 14 = 36 + 14\]So, \(12x\) must be equal to \(36\).
For the equation to stay balanced, \(x\) must be equal to \(3\):
\[12 \times 3 = 36\]In other words, \(x = 3\) makes the equation true:
\[12(3) + 14 = 36 + 14\]Worked Example 7.5: Finding the solution by inspection
What makes the following equations true?
- Find the number that goes in the box: \(\boxdot + \ 9 = - 7\)
- For this equation, solve for \(a\): \(a + 9 = - 7\)
Solve the equations.
Word it like this:
“Some number \(+ 9\) will give you \(-7\).”
In this case the number is \(-16\).
\[( - 16) + 9= - 7\]So, the value in the box is \(- 16\) and the value of \(a\) is \(- 16\).
Solve equations with fractions by inspection
When we solve equations, we are looking for the value of the unknown variable that will make the left-hand side equal to the right-hand side. We will solve these equations using inspection, which means looking carefully. We will look at the two sides of the equation, and try to match them up exactly.
Worked Example 7.6: Finding the solution by inspection
Which value of \(x\) makes the following equation true?
\[\frac{100}{x} = 20\]Inspect the two sides of the equation.
Left-hand side: \(\frac{100}{x}\) or \(100 \div x\)
The number \(100\) is divided by the variable \(x\).
Right-hand side: \(20\)
So, the division on the left-hand side must produce the answer \(20\).
We know that \(100 \div 5 = 20\).
Balance the equation.
\[100 \div x = 100 \div 5\]So, \(x\) must be equal to \(5\) to keep the equation balanced.
In other words, \(x = 5\) makes the equation true:
\[\frac{100}{5} = 20\]Worked Example 7.7: Solving equations with fractions
Solve for \(y\):
\[\frac{3}{5} = \frac{3}{y}\]Compare each side of the equation.
\[\frac{3}{5} = \frac{3}{y}\]We can see that the numerators are already the same. For both sides to be equal, the denominators must also be equal.
Find the unknown variable.
This question is “\(3\) over what is equal to \(3\) over \(5\)?” So \(y\) must be equal to \(5\).
\[\begin{align} \frac{3}{5} &= \frac{3}{y} \\ \therefore y &= 5 \end{align}\]We could also solve this equation by multiplying both sides by \(y\) and use inverse operations:
\[\begin{align} \frac{3}{5} \times y &= \frac{3}{y} \times y \\ \frac{3y}{5} &= 3 \\ 3y &= 3 \times 5 \\ 3y &= 15 \\ y &= 15 \div 3 \\ y &= 5 \end{align}\]We will work with this method in the next chapter. The method would also work if the numerators had not been the same.
Worked Example 7.8: Solving equations with fractions
Solve for \(y\):
\[\frac{1}{2y} = \frac{3}{18}\]Compare each side of the equation.
We should always first try to simplify each side of the equation. Here we can see that the fraction on the right can be simplified.
\[\begin{align} \frac{3}{18} &= \frac{3 \div 3}{18 \div 3} \\ &= \frac{1}{6} \end{align}\]Now the numerators are the same. For both sides to be equal, the denominators must also be equal.
Find the unknown variable.
\[\begin{align} \frac{1}{2y} &= \frac{3}{18} \\ \frac{1}{2y} &= \frac{1}{6}\\ 2y &= 6 \\ \therefore y &= 3 \end{align}\]Worked Example 7.9: Solving equations with fractions
Solve for \(b\):
\[- \frac{2}{4} = - \frac{1}{b + 5}\]Compare each side of the equation.
We should always first try to simplify each side of the equation. Here we can see that the fraction on the left can be simplified.
\[\begin{align} - \frac{2}{4} &= - \frac{2 \div 2}{4 \div 2} \\ &= - \frac{1}{2} \end{align}\]Now the numerators are the same. For both sides to be equal, the denominators must also be equal. Both sides also have the same sign, both are negative.
Find the unknown variable.
\[\begin{align} - \frac{2}{4} &= - \frac{1}{b + 5} \\ - \frac{1}{2} &= - \frac{1}{b + 5} \\ 2 &= b + 5 \\ \therefore b &= - 3 \end{align}\]Equations with exponents
One kind of exponential equation that you deal with in Grade 8 has one or more terms with a base that is raised to a power containing a variable.
For example: \(2^{x} = 16\).
When we need to find the unknown value, we are asking the question: “To what power must the base be raised for the statement to be true?”
- Make sure that the terms with \(x\) are on their own on one side: \(2^{x} = 16\)
- Write the known term in the same base as the term with the exponent: \(2^{x} = 2^{4}\)
- Equate the exponents: \(x = 4\)
In the example above, we can equate the exponents because the two numbers are equal only when they are raised to the same power.
In another kind of equation involving exponents, the variable is in the base.
For example: \(x^{2} = 25\).
When we need to find the unknown value, we are asking the question: “Which number must be raised to the given power for the statement to be true?”
For these equations, you should remember what you know about the powers of numbers such as \(2\), \(3\), \(4\), \(5\) and \(10\).
Here are some of the powers you will use often:
| \(\mathbf{x^{2}}\) | \(\mathbf{x^{3}}\) | \(\mathbf{x^{4}}\) | \(\mathbf{x^{5}}\) | |
| \(\mathbf{x = 1}\) | \(1^{2} = 1\) | \(1^{3} = 1\) | \(1^{4} = 1\) | \(1^{5} = 1\) |
| \(\mathbf{x = 2}\) | \(2^{2} = 4\) | \(2^{3} = 8\) | \(2^{4} = 16\) | \(2^{5} = 32\) |
| \(\mathbf{x = 3}\) | \(3^{2} = 9\) | \(3^{3} = 27\) | \(3^{4} = 81\) | \(3^{5} = 243\) |
| \(\mathbf{x = 4}\) | \(4^{2} = 16\) | \(4^{3} = 64\) | \(4^{4} = 256\) | \(4^{5} = 1\ 024\) |
| \(\mathbf{x = 5}\) | \(5^{2} = 25\) | \(5^{3} = 125\) | \(5^{4} = 625\) | \(5^{5} = 3\ 125\) |
Worked Example 7.10: Solving equations with exponents
Solve for \(x\):
\[8 = 2^{x}\]List the first few powers of \(2\).
To answer this question, we can list powers of \(2\), and see which one is equal to \(8\).
\[\begin{align} 2^{1} &= 2 \\ 2^{2} &= 4 \\ 2^{3} &= 8 \\ 2^{4} &= 16 \\ 2^{5} &= 32 \end{align}\]Find the power of \(x\) that matches the equation.
We can see that:
\[\begin{align} 8 &= 2^{3} \\ \therefore x &= 3 \end{align}\]Remember that the exponents tell us how many times to multiply the base by itself, so
\[\begin{align} 2^{3} &= 2 \times 2 \times 2 \\ &= 8 \end{align}\]
Worked Example 7.11: Solving equations with exponents
Solve for \(x\):
\[9^{x} = 9^{3}\]This equation has an \(x\) in the exponent, so it is not linear (it is exponential).
Find the question that the equation is asking.
When we solve equations, we are looking for the value of \(x\) that will make the left hand side equal to the right hand side. We will solve non-linear equations using inspection, which means looking carefully. We will look at each side and try to match them up exactly. This question is “\(9\) to the power of what is equal to \(9\) to the power of \(3\)?”.
\[9^{x} = 9^{3}\]Find the value of the variable (if the bases are the same).
We can see that the bases are already the same. For both sides to be equal, the exponents must also be equal.
\[\begin{align} 9^{x} &= 9^{3} \\ \therefore x &= 3 \end{align}\]Both sides are equal to \(9^{3} = 729\). So, we have found the \(x\)-value that makes the equation true.
Worked Example 7.12: Solving equations with exponents
Solve for \(x\):
\[x^{4} = 625\]Find the question that the equation is asking.
To answer this question, look at the right-hand side of the equation. The question is “What number raised to the power of \(4\) will give \(625\)?”.
Use your maths experience to find what will answer this question.
The number \(625\) ends in a \(5\), so it is likely that it is a multiple of \(5\). Try the number \(5\):
\[\begin{align} 625 &= 5 \times 5 \times 5 \times 5 \\ 625 &= 5^{4} \end{align}\]Find the value of the variable (if the exponents are the same).
We now have both sides with the same power:
\[\begin{align} x^{4} &= 5^{4} \\ \therefore x &= 5 \end{align}\]Worked Example 7.13: Solving equations with exponents
Solve for \(x\):
\[x^{3} = - 64\]Inspect the equation.
Let’s think about \(x^{3} = - 64\), and remember that it means the same as \(x \times x \times x = - 64\). We are being asked “What number, when you cube it, gives \(- 64\)?”.
Use known facts to find the solution.
To help us answer this question, we will list the first few positive cubes and see which one is equal to \(64\):
\[\begin{align} 1^{3} &= 1 \\ 2^{3} &= 8 \\ 3^{3} &= 27 \\ 4^{3} &= 64 \end{align}\]By inspection, we can see that \(x = 4\), since we know that \(4 \times 4 \times 4 = 64\).
Now, because we have a negative number on the right-hand side, the number we multiply by itself (or raise to the power of \(3\)) must also be negative:
\[- 64 = {( - 4)}^{3}\]We now have both sides with the same power:
\[\begin{align} x^{3} &= ( - 4)^{3} \\ \therefore x &= - 4 \end{align}\]Equations with roots
To solve the equation, we must find the value of the variable that will make both sides of the equation equal. When the variable is inside a square root sign (sometimes called a radical), we ask the question, “What number, when you take its square root, is equal to the given number?”
For example, \(7 = \sqrt{b}\). We ask: “What number, when you take its square root, is equal to \(7\)?”
The number in this case is \(49\), because \(\sqrt{49} = 7\).
Worked Example 7.14: Solving equations with roots
Solve for \(n\):
\[\sqrt{n} = 13\]Find the number that gives \(13\) when you take its square root.
To solve the equation, we must find the value of \(n\) that will make both sides of the equation equal. The \(n\) is inside a square root sign. So, ask the question, “What number, when you take its square root, is equal to \(13\)?”
It is true that:
\[\sqrt{169} = 13\]Solve for \(n\) by inspection.
Since \(\sqrt{n} = 13\), and it is true that \(\sqrt{169} = 13\):
\[n = 169\]Check your solution.
You should always check your answer:
- LHS: \(n = \sqrt{169} = 13\)
- RHS: \(13\)
Since LHS = RHS, our solution is correct.
\[\therefore n = 169\]Worked Example 7.15: Solving equations with roots
Solve for \(m\):
\[\sqrt{8} = \sqrt{m + 13}\]Solve for \(m\) by inspection.
To solve this equation, we are looking for the value of \(m\) that will make the left-hand side of the equation equal to the right-hand side. Both sides already have square roots, so the parts underneath the square roots must be equal.
\[\begin{align} \sqrt{8} &= \sqrt{m + 13} \\ \therefore 8 &= m + 13 \end{align}\]By inspection, we can see that \(- 5 = m\).
Check your solution.
We can only take the square root of a positive value. We should check our answer.
- LHS: \(\sqrt{m + 13} = \sqrt{- 5 + 13} = \sqrt{8}\)
- RHS: \(\sqrt{8}\)
Since LHS = RHS, our solution is correct.
\[\therefore m = - 5\]Equations with powers and brackets
We name equations by looking at the highest power of \(x\) in the equation:
- Equations with \(x\) terms only are called linear. For example, \(y = 2x + 1\).
- Equations with \(x^{2}\) terms are called quadratic. For example, \(y = x^{2}\).
- Equations with \(x^{3}\) terms are called cubic. For example, \(y = 2x^{3}\).
Worked Example 7.16: Solving quadratic equations
Solve for \(b\):
\[b^{2} = 169\]Solve by inspection.
This equation has an \(b^{2}\) term, so it is not linear (it is quadratic).
When we solve equations, we are looking for the value of \(b\) that will make the left-hand side equal to the right-hand side. We will solve non-linear equations using inspection. This question is asking “What number (or numbers), when you square it, gives \(169\)?”
Since \(13 \times 13 = 169\), we know that \(13\) is a solution.
Check for other solutions.
However, this is not the only solution. Remember, when you square a negative number you get a positive number.
\[\begin{align} ( - 13)^{2} &= ( - 13)( - 13) \\ &= 169 \end{align}\]Therefore, \(−13\) is also a solution. Either of the two answers will make the equation true, so we write
\[x = 13 \text{ or } x = - 13\]Equations can have more than one solution – this equation has two answers. Both values make the equation true, so we can’t choose between them. Instead we give both answers.
Worked Example 7.17: Solving cubic equations
Solve for \(b\):
\[(b - 3)^{3} = 64\]Find the number that gives \(64\) when cubed.
When we solve equations, we are looking for the value of the variable that will make the left-hand side equal to the right-hand side. This equation has a \(b^{3}\) term and so it is not linear, but cubic. For now, we solve non-linear equations by inspection.
First, we need to work out which number gives \(64\) when cubed.
\[(b - 3)^{3} = 64\]To help answer this question, we will list the first few positive cubes and see which one is equal to \(64\):
\[\begin{align} 1^{3} &= 1 \\ 2^{3} &= 8 \\ 3^{3} &= 27 \\ 4^{3} &= 64 \end{align}\]So, it follows that \((b - 3) = 4\).
Solve for \(b\).
We write:
\[\begin{align} (b - 3)^{3} &= 64 \\ (b - 3)^{3} &= 4^{3} \\ b - 3 &= 4 \\ b &= 7 \end{align}\]When we set the bases equal to each other, we reduced the equation to a linear equation, which we can also solve using inverse operations.