Tables and Values
Every question with a fully worked solution.
- Tables and Values - Video - Completing a table of values Watch
Theory
A table of values holds the \(y\)-value that a rule gives for each \(x\)-value. Work one column at a time: for \(y = 2x + 3\), \(x = 0, 1, 2\) gives \(y = 3, 5, 7\). This Year 8 maths page shows how, with negatives, fractions and missing \(x\)-values.
Rule to table. Substitute each \(x\)-value, following the order of operations: multiply or divide first, then add or take away. For \(y = 3x + 1\) at \(x = -2\): \(-6 + 1 = -5\).
Subtraction written first. In \(y = 8 - 5x\), work out \(5x\), then take it from \(8\). At \(x = -1\): \(8 - (-5) = 8 + 5 = 13\).
A fraction bar is a bracket. In \(y = \dfrac{x + 6}{3}\), add \(6\) first, then divide by \(3\). If it does not divide exactly, leave a fraction.
No rule given? Find the change from one \(y\)-value to the next. If the \(x\)-values go up in ones, keep adding that change to fill the gaps. For a far column, write the rule first, then substitute.
Uneven \(x\)-values. Find the change in \(y\) for one step in \(x\), then multiply by the size of the jump.
Working backwards. If a \(y\)-value is known and its \(x\)-value is not, put the \(y\)-value into the rule and solve.
Check. When \(x\) goes up in ones, the \(y\)-row changes by the same amount each time. A column that breaks the run is wrong. Each column is also a point \((x, y)\) to plot.
What to do first:
| Rule | Do first | Then |
|---|---|---|
| \(y = 4x - 1\) | \(4 \times x\) | take away \(1\) |
| \(y = 12 - 5x\) | \(5 \times x\) | take it from \(12\) |
| \(y = \dfrac{x + 6}{3}\) | \(x + 6\) | divide by \(3\) |
| \(y = \dfrac{x}{5} + 2\) | \(x \div 5\) | add \(2\) |
Completing a table of values
- Read the \(x\)-row first; note any negatives, decimals or jumps.
- Substitute each \(x\)-value into the rule, one column at a time.
- Follow the order of operations.
- Check that the \(y\)-values change by a constant amount for each step in \(x\).
| \(x\) | \(0\) | \(1\) | \(2\) |
|---|---|---|---|
| \(y\) | \(a\) | \(b\) | \(c\) |
| \(a\) | \(=\) | \(2 \times 0 + 3 = 3\) |
| \(b\) | \(=\) | \(2 \times 1 + 3 = 5\) |
| \(c\) | \(=\) | \(2 \times 2 + 3 = 7\) |
\(a = 3\), \(b = 5\), \(c = 7\)
| \(x\) | \(0\) | \(1\) | \(2\) |
|---|---|---|---|
| \(y\) | \(a\) | \(b\) | \(c\) |
| \(a\) | \(=\) | \(0 + 2 = 2\) |
| \(b\) | \(=\) | \(-5 + 2 = -3\) |
| \(c\) | \(=\) | \(-10 + 2 = -8\) |
\(a = 2\), \(b = -3\), \(c = -8\)
| \(x\) | \(-2\) | \(0\) | \(1.5\) | \(4\) |
|---|---|---|---|---|
| \(y\) | \(a\) | \(-4\) | \(b\) | \(c\) |
| \(3 \times (-2) - 4\) | \(=\) | \(-10\) |
| \(3 \times 1.5 - 4\) | \(=\) | \(0.5\) |
| \(3 \times 4 - 4\) | \(=\) | \(8\) |
\(a = -10\), \(b = 0.5\), \(c = 8\). Check the given column: \(0 - 4 = -4\).
| \(x\) | \(-3\) | \(0.5\) | \(4\) |
|---|---|---|---|
| \(y\) | \(a\) | \(b\) | \(c\) |
| \(7 - (-6)\) | \(=\) | \(13\) |
| \(7 - 1\) | \(=\) | \(6\) |
| \(7 - 8\) | \(=\) | \(-1\) |
\(a = 13\), \(b = 6\), \(c = -1\)
(ii) Find \(y\) when \(x = 9\).
| \(x\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) |
|---|---|---|---|---|---|
| \(y\) | \(11\) | \(2\) | \(-7\) | \(a\) | \(b\) |
The change is \(2 - 11 = -9\) each time. For \(x = 9\), write the rule: start \(11\), change \(-9\).
| \(a\) | \(=\) | \(-7 - 9 = -16\) |
| \(b\) | \(=\) | \(-16 - 9 = -25\) |
| \(y\) | \(=\) | \(11 - 9x\) |
| \(11 - 9 \times 9\) | \(=\) | \(-70\) |
(i) \(a = -16\), \(b = -25\)
(ii) \(-70\)
| \(x\) | \(0\) | \(2\) | \(12\) |
|---|---|---|---|
| \(y\) | \(a\) | \(b\) | \(c\) |
| \(\dfrac{0 + 6}{3}\) | \(=\) | \(2\) |
| \(\dfrac{2 + 6}{3}\) | \(=\) | \(\dfrac{8}{3}\) |
| \(\dfrac{12 + 6}{3}\) | \(=\) | \(6\) |
\(a = 2\), \(b = \dfrac{8}{3}\), \(c = 6\). Every \(3\) in \(x\) adds \(1\) to \(y\).
(i) Find \(V\) when \(d = 2\) and \(d = 5\).
(ii) On which day is there \(300\) L left?
| \(900 - 60 \times 2\) | \(=\) | \(780\) |
| \(900 - 60 \times 5\) | \(=\) | \(600\) |
| \(60d\) | \(=\) | \(900 - 300 = 600\) |
| \(d\) | \(=\) | \(10\) |
(i) \(780\) L and \(600\) L
(ii) Day \(10\)
(ii) For \(y = 5x + 2\), find \(x\) when \(y = 37\) and when \(y = -8\).
(i) From \(x = 0\) to \(x = 3\), \(y\) goes up \(6\), so \(2\) for each \(1\) in \(x\).
| \(a\) | \(=\) | \(2 + 2 \times 7 = 16\) |
| \(y\) | \(=\) | \(2 + 2 \times (-2) = -2\) |
| \(5x + 2\) | \(=\) | \(37\) |
| \(x\) | \(=\) | \(7\) |
| \(5x + 2\) | \(=\) | \(-8\) |
| \(x\) | \(=\) | \(-2\) |
(i) \(a = 16\); \(y = -2\) when \(x = -2\)
(ii) \(x = 7\) and \(x = -2\)
Common pitfalls
Frequently asked questions
How do I complete a table of values from a rule?
Substitute each x-value into the rule, one column at a time, using the order of operations.
What if the x-value is negative?
Use brackets when you substitute, such as 3 × (−2) = −6.
How do I fill a table that has no rule?
Find the change from one y-value to the next and keep adding it, or write the rule and substitute.
How do I find a missing x-value?
Put the y-value into the rule and solve the equation for x.
How can I check my table?
If x goes up in ones, the y-values must change by the same amount each time.