Inequalities and simultaneous equations: a week of ready-to-teach maths
Five lessons for Year 10 Algebra. Print this pack and the week is prepared: each day has a one-page plan and a student worksheet with practice and a stretch section, plus twelve number-line cards, twelve equation-pair cards, a mini-check and every answer with its working.
Start here: five minutes to Monday
- Skim the week at a glance on the next page.
- Print the five days. Each day is two A4 sheets: a plan and a worksheet.
- Print the two cut-out sheets, one copy per pair. Cut the 12 number-line cards before Day 2 and the 12 equation-pair cards before Day 4, and keep each set in an envelope.
- Students need a ruler on Day 3 for the graphs; the grids are printed on the worksheet. Calculators are optional.
- Open the free interactive unit on the board. Every plan says which picture to show and which buttons to press.
The idea in plain words
An equation pins down a value; an inequality allows a whole range of them. Solving an inequality works like solving an equation, with one difference: multiplying or dividing both sides by a negative number reverses the sign, because it reverses the order of numbers on the line. Two equations in x and y are two lines, and a pair of values that satisfies both is a point on both lines: where they cross. Algebra finds that point exactly, by substitution or elimination, and the answer only means something once it is said back in the situation: 80 adult tickets, or 100 minutes.
One day, one lesson
The five lessons fill a week of maths, one lesson of about 60 minutes a day. Run them in order: each day stands on the one before. Every plan marks a split point for two shorter sessions, and every worksheet ends with a stretch section for students who finish early.
How we write these
We use < and > when the boundary is not included, and ≤ and ≥ when it is. On a number line an open circle marks a boundary that is left out and a filled circle one that is included. A solution to a pair of equations is written x = 2, y = 5 or as the point (2, 5), and every answer to a problem ends with a sentence about the situation.
The week at a glance
One lesson a day for a week. Each day stands on the day before, so run them in order.
| Day | Lesson | Students learn and do | On screen |
|---|---|---|---|
| 1 | An answer that is a range | Solve linear inequalities, show each answer on a number line with an open or a filled circle, test values, and turn two situations into inequalities | An inequality's answer is a range |
| 2 | When the sign flips | See with numbers why multiplying or dividing by a negative reverses the sign, solve with letters on both sides, and match twelve number-line cards | Dividing by a negative flips the sign |
| 3 | Where two lines cross | Graph two lines from tables, read the crossing point as the solution, meet parallel lines and a repeated line, and compare two phone plans | Simultaneous equations: where two lines cross |
| 4 | Substitution and elimination | Solve pairs of equations by substitution and by elimination, check both equations, and match twelve cards with swapped twins among the answers | Solving by substitution |
| 5 | Regions and real situations | Decide which side of a line to shade with a test point, solve a ticket problem, test a shopping budget, and play the self-check quiz as a class game | An inequality in two variables shades a region; self-check quiz as a class game |
How the week builds
Days 1 and 2 solve inequalities in one letter; Day 3 shows a pair of equations as two lines; Day 4 finds the crossing point by algebra; and Day 5 brings both together, as regions on a graph and as answers to real questions. It builds on Year 9, where students graphed linear relations and found gradients and intercepts, and on the linear equations of Year 8. It prepares the modelling of growth and decay later in Year 10, where answers are again read back into a situation.
Materials for the week (one trip)
- From the classroom: rulers, scissors, envelopes for the cards, pencils, this pack printed.
- Cut-out sheet 1, one per pair: 12 number-line cards (Day 2).
- Cut-out sheet 2, one per pair: 12 equation-pair cards (Day 4).
- If you have them: graphing software or a graphics calculator to check the Day 3 graphs. No other equipment to buy.
Words for the word wall
inequality, less than, greater than, at most, at least, boundary, open circle, closed circle, solution set, simultaneous equations, intersection, substitution, elimination, parallel, region, half-plane, test point
Dear families
This week in maths, Year 10 is solving inequalities, where the answer is a range of values, and pairs of equations, where two conditions must be true at once. Both turn up whenever a limit or a choice is involved: a budget, a lift’s weight limit, or two phone plans.
Try this at home
- Ask: a lift holds at most 900 kg. Which sign fits, < or ≤? (≤: exactly 900 kg is allowed.)
- Ask: is −5 bigger or smaller than −2? (Smaller: it is further left on the number line.)
- Ask: two numbers add to 10 and differ by 4. What are they? (7 and 3.)
- Ask: one plan costs $20 plus 10 cents a minute, another $5 plus 25 cents a minute. When do they cost the same? (100 minutes.)
My limits week
Fill one row a day. Write something with a limit, the inequality, and whether the limit itself is allowed.
| Day | Something with a limit | The inequality | Limit allowed? |
|---|---|---|---|
| Monday | |||
| Tuesday | |||
| Wednesday | |||
| Thursday | |||
| Friday |
An answer that is a range
An inequality is solved like an equation, but its answer is a range of values, drawn as a ray on a number line. Students solve four inequalities, test values against one, and write two situations as inequalities with the answer in words.
We are learning to
- solve a linear inequality by doing the same to both sides,
- show the solution on a number line with an open or a filled circle,
- write a situation as an inequality and give the answer in words.
Success criteria
- I can solve 2x + 5 < 1 and draw the answer on a number line.
- I can say what b ≤ 27 means for a lift that carries boxes.
You need
The worksheet, one per student. Calculators are optional.
Lesson flow (about 60 minutes)
| 12 min | A range of answers Show the board (see the box), then solve the first row of Part A together and draw it. Ask: “Is 3 a solution of x ≥ 3? Of x > 3? How does the circle show the difference?” |
| 25 min | Solve and test Pairs complete Parts A and B. Ask: “Which value in Part B sits right on the boundary, and does it count?” |
| 23 min | Situations Pairs write and solve Part C, ending each with a sentence; early finishers do Part D. |
Two shorter sessions? Stop after Part B; do the situations and Part D next time.
Watch for these ideas
- “The answer is x = 5.” x > 5 is every number bigger than 5, a whole range.
- Mixing the circles: open leaves the boundary out (< or >), filled includes it (≤ or ≥).
- Ignoring the situation: b ≤ 27 means 0 to 27 whole boxes, not 27.5 and not −3.
Answers
- Part A: x + 4 > 9 gives x > 5 (open circle at 5, ray to the right); 3x ≤ 12 gives x ≤ 4 (filled circle at 4, ray to the left); x/2 − 1 ≥ 2 gives x ≥ 6 (filled circle at 6, ray to the right); 2x + 5 < 1 gives x < −2 (open circle at −2, ray to the left).
- Part B: x > 2, so 4, 6, 2.5 work and 2, 0, −1 do not; 2 is the boundary and gives 5 > 5, which is false.
- Part C: 90 + 30b ≤ 900 gives b ≤ 27: at most 27 boxes. 15 + 0.2m < 25 gives m < 50: up to 49 whole minutes.
- Part D: subtract 1 and halve: 1 < x ≤ 4, an open circle at 1, a filled one at 4 and the segment between them shaded.
An answer that is a range
A. Solve and draw
Solve each inequality, then draw the answer on its number line. The first one is done for you.
| 1. x + 4 > 9 so x > 5 | 2. 3x ≤ 12 |
| 3. x/2 − 1 ≥ 2 | 4. 2x + 5 < 1 |
B. Test the values
Circle every value that makes 3x − 1 > 5 true: 2, 4, 0, 6, 2.5, −1
C. Situations
D. Stretch: two at once
When the sign flips
Multiplying or dividing both sides by a negative number reverses the order of numbers on the line, so the inequality sign must flip. Students see it with numbers first, solve inequalities that need the flip and then letters on both sides, and match twelve number-line cards.
We are learning to
- explain with numbers why multiplying or dividing by a negative reverses an inequality,
- solve inequalities that need the flip, and check each with a test value,
- solve inequalities with the letter on both sides.
Success criteria
- I can solve 7 − 3x ≥ 1 and check my answer with a test value.
- I can explain why −2x < 6 gives x > −3, not x < −3.
You need
The number-line cards (cut-out sheet 1), one set per pair; the worksheet, one per student.
Lesson flow (about 60 minutes)
| 12 min | Numbers first Fill the first row of the Part A table together, then show the board (see the box). Ask: “Multiply both sides of 2 < 5 by −1. Which is bigger now, −2 or −5?” |
| 28 min | Flip and check Pairs finish Parts A and B, testing every answer with one value. Ask: “Could you avoid dividing by a negative by moving the x term to the other side?” |
| 20 min | Match the cards Pairs match the twelve cards (Part C) and say which pairs needed a flip; early finishers do Part D. |
Two shorter sessions? Stop after Part A; start the next with Part B.
Watch for these ideas
- Flipping after adding or subtracting: only multiplying or dividing by a negative flips the sign.
- Flipping because the other side is negative: 3x < −6 gives x < −2; dividing by +3 keeps the sign.
- Skipping the check: one test value catches a missed flip in seconds.
Answers
- Part A: 6 < 15 (× 3), −2 > −5 (× (−1)), −8 > −20 (× (−4)), −1 > −2.5 (÷ (−2)); −3x > 12 gives x < −4; −x/2 ≤ 3 gives x ≥ −6; 7 − 3x ≥ 1 gives x ≤ 2; 5 − 2x < 11 gives x > −3.
- Part B: 5x − 2 > 3x + 6 gives x > 4; 2(x + 3) ≤ 5x − 9 gives x ≥ 5; 3 − x ≥ x − 9 gives x ≤ 6.
- Part C: the card key is with the mini-check answers.
- Part D: 2 < 4: 1/2 > 1/4; −4 < −2: −1/4 > −1/2; −1 < 2: −1 < 1/2. It holds when a and b have the same sign, but not across zero.
When the sign flips
A. The flip
Start from 2 < 5 each time. Apply the step to both sides and write < or > between the results. The first row is done for you.
| Step | Left | < or > | Right |
|---|---|---|---|
| × 3 | 6 | < | 15 |
| × (−1) | |||
| × (−4) | |||
| ÷ (−2) |
Now solve, and check each answer with a test value.
| 1. −3x > 12 | 2. −x/2 ≤ 3 |
| 3. 7 − 3x ≥ 1 | 4. 5 − 2x < 11 |
B. Letters on both sides
C. Match the cards
Match each inequality card to the number line of its solution. Write the letters, and mark the pairs that needed a flip.
| Pair 1: ______ and ______ flip? ____ | Pair 2: ______ and ______ flip? ____ |
| Pair 3: ______ and ______ flip? ____ | Pair 4: ______ and ______ flip? ____ |
| Pair 5: ______ and ______ flip? ____ | Pair 6: ______ and ______ flip? ____ |
D. Stretch: upside down
Where two lines cross
Two equations in x and y are two lines, and their simultaneous solution is the one point on both. Students tabulate and graph two pairs of lines, read the crossing points, meet parallel lines and a repeated line, and compare two phone plans.
We are learning to
- graph a linear equation from a table of values,
- read the solution of two equations as the crossing point of their graphs,
- recognise when two equations have no solution, or infinitely many.
Success criteria
- I can graph y = 2x + 1 and y = 7 − x and read their solution.
- I can explain from the equations why y = 2x + 1 and y = 2x − 3 never meet.
You need
Rulers; the worksheet, one per student; graphing software if you have it, to check the graphs.
Lesson flow (about 60 minutes)
| 12 min | Two lines Show the board (see the box), then fill both rows of the Part A table together. Ask: “Where do the two rows of y agree? What does that column tell us?” |
| 30 min | Graph and read Pairs graph Parts A and B, read each crossing point and check it in both equations. Ask: “Why must the crossing point satisfy both equations?” |
| 18 min | Special pairs Pairs reason through Part C from the equations alone; early finishers do Part D. |
Two shorter sessions? Stop after Part A; do Part B, the special pairs and Part D next time.
Watch for these ideas
- Reading only one coordinate: the solution is a pair, x and y together.
- Plotting y = 7 − x as rising: the −x makes it fall 1 for each 1 across.
- “Parallel lines meet far away.” Same gradient, different intercepts: they never meet, so there is no solution.
Answers
- Part A: y = 2x + 1: 1, 3, 5, 7, 9; y = 7 − x: 7, 6, 5, 4, 3. They agree at x = 2: x = 2, y = 5.
- Part B: y = x + 3: 3, 4, 5, 6, 7; y = 9 − x: 9, 8, 7, 6, 5. x = 3, y = 6.
- Part C: 1. no solution: the lines are parallel: same gradient, 2, but different intercepts. 2. every point of the line: the two equations are the same line: halve 2y = 2x + 2 to get y = x + 1.
- Part D: C = 20 + 0.1m and C = 5 + 0.25m meet at m = 100, C = 30: the plans cost the same, $30, at 100 minutes. At 150 minutes Plan A costs $35.00 and Plan B $42.50, so Plan A is cheaper above 100 minutes and Plan B below.
Where two lines cross
A and B. Tables, then graphs
Fill each table, plot both lines on the grid below it, and read where they cross. Check the point in both equations.
| x | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| y = 2x + 1 | |||||
| y = 7 − x |
| x | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| y = x + 3 | |||||
| y = 9 − x |
C. Special pairs
Without graphing, decide how many solutions each pair has, and explain.
D. Stretch: two phone plans
Substitution and elimination
Algebra finds the crossing point exactly. Substitution replaces one letter using an equation that gives it; elimination adds or subtracts the equations so that one letter cancels. Students solve pairs both ways and match twelve cards with swapped twins among the answers.
We are learning to
- solve a pair of equations by substitution,
- solve a pair of equations by elimination, adding or subtracting to cancel a letter,
- check a solution in both equations, and choose the quicker method.
Success criteria
- I can solve y = 3x − 2 and x + y = 10 by substitution.
- I can solve 2x + 3y = 13 and 2x − y = 1 by elimination and check both equations.
You need
The equation-pair cards (cut-out sheet 2), one set per pair; the worksheet, one per student.
Lesson flow (about 60 minutes)
| 12 min | Swap it in Show the board (see the box), then solve the first pair of Part A together. Ask: “Why are we allowed to set x + 1 equal to 5 − x?” |
| 28 min | Cancel a letter Pairs finish Part A, then solve Part B by elimination. Ask: “When do you add the two equations, and when do you subtract?” |
| 20 min | Match and check Pairs match the twelve cards (Part C), checking each answer in both equations; early finishers do Part D. |
Two shorter sessions? Stop after Part A; start the next with Part B.
Watch for these ideas
- Checking one equation only: (3, 2) fits x + y = 5 but not y = 2x − 1. The cards are built to catch this.
- Subtracting only some terms: when one equation is taken from another, every term changes sign, the constants too.
- Stopping at x: the answer is a pair, so substitute back to find y.
Answers
- Part A: y = 3x − 2 and x + y = 10: x = 3, y = 7; y = 2x and 3x + y = 20: x = 4, y = 8.
- Part B: x + y = 9 and x − y = 3: x = 6, y = 3; 2x + 3y = 13 and 2x − y = 1: x = 2, y = 3; 3x + 2y = 16 and x + 2y = 8: x = 4, y = 2.
- Part C: the card key is with the mini-check answers.
- Part D: multiply 3x + 2y = 12 by 3 and 2x − 3y = 8 by 2, so the y terms are +6y and −6y; adding cancels y. x = 4, y = 0.
Substitution and elimination
A. Substitution
One equation already gives y. Put it into the other, solve for x, then find y. Check in both equations.
B. Elimination
Add or subtract the equations so that one letter cancels. Solve, then substitute back. Check in both equations.
C. Match the cards
Match each pair of equations to its solution. Watch for swapped twins such as (2, 3) and (3, 2): check both equations.
| Pair 1: ______ and ______ | Pair 2: ______ and ______ |
| Pair 3: ______ and ______ | Pair 4: ______ and ______ |
| Pair 5: ______ and ______ | Pair 6: ______ and ______ |
D. Stretch: multiply first
Regions and real situations
An inequality in x and y is solved by a region: every point on one side of a boundary line. A test point decides the side. Students decide the side for four inequalities, solve a ticket problem, test a shopping budget, and play the self-check quiz.
We are learning to
- decide which side of a boundary line to shade, using a test point,
- write and solve a pair of equations from a situation,
- give each answer as a sentence about the situation.
Success criteria
- I can say whether (0, 0) satisfies y ≥ 2x − 1, and which side to shade.
- I can solve the ticket problem and say how many tickets of each kind were sold.
You need
The worksheet, one per student; the unit’s self-check quiz on the board for the game.
Lesson flow (about 60 minutes)
| 12 min | Which side? Show the board (see the box), then do the first row of Part A together. Ask: “Why is (0, 0) a handy test point? When can it not be used?” |
| 30 min | Situations Pairs complete Parts A to C, ending each answer with a sentence; early finishers do Part D. Ask: “Does “at most $60” include $60 exactly? What does that say about the boundary line?” |
| 18 min | Self-check quiz as a class game Run the unit’s self-check quiz on the board. Teams vote on each answer, then the reason is read aloud. |
Two shorter sessions? Stop after Part B; do the budget, Part D and the quiz next time.
Watch for these ideas
- Testing a point on the line: it gives equality, so it cannot decide the side. Pick one off the line.
- Shading by the look of the sign: x + 2y ≤ −2 is not obviously above or below. Use a test point.
- Answers without words: a = 80 is not finished; 80 adult tickets and 120 student tickets is.
Answers
- Part A: y ≥ 2x − 1: yes, shade the side with the origin; y < 4 − x: yes, shade the side with the origin; y > x + 2: no, shade the side away from the origin; x + 2y ≤ −2: no, shade the side away from the origin.
- Part B: a + s = 200 and 12a + 5s = 1560; substitute s = 200 − a: 12a + 5(200 − a) = 1560, so 7a = 560, a = 80 and s = 120: 80 adult tickets and 120 student tickets.
- Part C: 10 L and 10 loaves cost $60: yes; 12 L and 10 loaves cost $64: no; 8 L and 9 loaves cost $52: yes. A cost of exactly $60 is on the boundary line and still allowed.
- Part D: 10 points: (0, 0), (0, 1), (0, 2), (0, 3), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (3, 0).
Regions and real situations
A. Which side?
Test the point (0, 0) in each inequality. Then say whether to shade the side of the line with the origin or the side away from it.
| Inequality | Is (0, 0) in the region? | Shade with or away from the origin? |
|---|---|---|
| y ≥ 2x − 1 | 0 ≥ −1, yes | with |
| y < 4 − x | ||
| y > x + 2 | ||
| x + 2y ≤ −2 |
B. Tickets
C. The shopping budget
A café may spend at most $60 on milk at $2 a litre and bread at $4 a loaf: 2m + 4b ≤ 60. Can it buy each of these?
| Milk (L) | Loaves | Cost ($) | Within budget? |
|---|---|---|---|
| 10 | 10 | ||
| 12 | 10 | ||
| 8 | 9 |
D. Stretch: points in a corner
Inequalities and their number lines
For Day 2, one set per pair. Six cards show an inequality; six show the solution of one on a number line. Solve each inequality and find its number line. Three of them need the sign to flip.
Teacher note: the key is with the mini-check answers.
Pairs of equations and their solutions
For Day 4, one set per pair. Six cards show a pair of equations; six show a point. Solve each pair and find its point, checking it in both equations: some points are swapped twins of each other.
3x − y = 7
x − y = 3
2x + y = 0
y = x + 3
x + y = 5
x + y = 5
Teacher note: the key is with the mini-check answers.
What we know: Inequalities and simultaneous equations
Work on your own. A calculator is allowed. Show your working.
- Solve 3x − 4 > 11 and draw the answer on the number line.
- Solve 6 − 2x ≤ 14. ______________________
- Circle every value that makes 2x + 3 ≥ 9 true: 1, 3, 4.5, −3, 6, 2.9
- Solve by substitution: y = x + 4 and 2x + y = 13. ______________________
- Solve by elimination: 3x + y = 14 and x − y = 2. ______________________
- How many solutions do y = 3x + 2 and y = 3x − 1 have? Circle one: one · none · infinitely many
- Does (0, 0) lie in the region y < 2x + 3? Which side of the line is shaded? ______________________
- A gym charges $40 to join and $10 a visit; another charges $25 a visit and nothing to join. Write an inequality and find how many visits make the first gym cheaper. ______________________
How did this week feel? Circle one: I can teach it · I am getting there · I need more help
Answers and marking guide
Answers with working
- x > 5: an open circle at 5 and a ray to the right.
- x ≥ −4: dividing by −2 flips the sign.
- x ≥ 3, so circle 3, 4.5, 6.
- x = 3, y = 7: x + 4 replaces y, so 3x + 4 = 13.
- x = 4, y = 2: adding cancels y, so 4x = 16.
- none: the gradients are the same, 3, and the intercepts differ, so the lines are parallel.
- Yes: 0 < 3 is true, so the side of the line with the origin is shaded (below the line).
- 40 + 10v < 25v, so 40 < 15v and v > 40/15 ≈ 2.67: the first gym is cheaper from 3 visits on (at 3 visits, $70 against $75).
A quick three-level guide
| Idea | Working towards | At standard | Beyond |
|---|---|---|---|
| Solving inequalities | Treats the answer as one number, or forgets the flip | Solves, flips when dividing by a negative, and draws the right circle | Explains the flip with numbers and checks with a test value |
| Graphs of two lines | Reads one coordinate only | Graphs both lines and reads the crossing point | Explains parallel and repeated lines from the equations alone |
| Solving by algebra | Checks one equation only | Solves by substitution and by elimination, checking both equations | Chooses the quicker method and says why |
| Situations | Leaves the answer as a letter | Writes the equations or inequality and answers in a sentence | Notices boundaries and whole-number limits in the situation |
Eight questions, four ideas. A student at standard answers most questions with working, checks pairs in both equations, and ends Question 8 with a sentence.
Card keys
Number-line cards (Day 2): G and A (x > 3), D and J (x ≥ 3), E and L (x < 3), B and H (x ≤ 3), K and F (x > −3), I and C (x ≤ −3). Equation-pair cards (Day 4): W and R (2, 3), U and O (3, 2), S and M (1, 4), P and V (4, 1), Q and X (−1, 2), N and T (2, −1).
Weekly class record
Jot a tick as you move around the room; the mini-check fills any gaps. A tick a day is plenty.
| Name | Ranges | The flip | Graphs | Algebra | Situations |
|---|---|---|---|---|---|
The five columns are the five days: answers that are ranges, the sign flip, graphs of two lines, solving by algebra, and regions and real situations.