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- 1Identify the gradient and y-intercept components in a linear relationship.
- 2Show learners a photograph of a steep hill (Kwame climbing Kwahu Mountain) and a gentle slope (a ramp at Makola Market). Ask: Which climb is steeper? How would you measure the steepness? Learners discuss with their partner and write one word describing steepness in their exercise books.
- 3Draw a simple right-angled triangle on the board with horizontal distance 4 units and vertical height 2 units. Ask: If you walk 4 steps forward and rise 2 steps up, what is the ratio of up to forward? Learners calculate and share the ratio aloud.
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- UNDERSTANDING GRADIENT AS RATE OF CHANGE
- 1Display the table of values: Distance walked by Ama (hours: 0, 1, 2, 3; kilometers: 0, 3, 6, 9). Ask learners to describe the pattern: For every 1 hour, how many kilometers? Write on the board: gradient = change in y / change in x = (6 − 3) / (2 − 1) = 3. Learners copy into exercise books and calculate gradient using two other pairs of points from the table.
- 2Using the ruler and graph board provided, learners plot the four ordered pairs (0,0), (1,3), (2,6), (3,9) for Ama's walk. They draw a straight line through the points and measure the vertical rise and horizontal run using the ruler. Compare the ratio they measure with the calculated gradient of 3.
- 3Struggling learners: work with the first two points only to find gradient. Fast finishers: predict the distance at 5 hours and verify by extending the line on the graph board.
- 4Emphasise that gradient is the steepness — the same concept as the hill example in Phase 1.
- WRITING LINEAR EQUATIONS FROM GRADIENT
- 5Display gradient = 3 and the point (0, 9) on the board. Explain: y = mx + c, where m is gradient and c is the y-intercept (where the line crosses the y-axis). Since the line crosses at 9, m = 3 and c = 9, so the equation is y = 3x + 9. Ask: What does this equation tell us about Kwaku's journey (if he starts 9 km away from home and travels 3 km per hour)? Learners write the equation in their exercise books.
- 6Provide learners with three more linear graphs drawn on graph boards showing: (i) gradient 2, y-intercept 5; (ii) gradient −1, y-intercept 4; (iii) gradient 0.5, y-intercept 2. Using the textbook examples as reference, learners write the equation y = mx + c for each graph.
- 7Struggling learners: use the formula card with y = mx + c printed and fill in m and c values. Fast finishers: create their own linear equation using a gradient between 2 and 4 and y-intercept between 0 and 10, then verify by plotting five points.
- 8Reference the textbook examples of building roofs and mountain slopes to contextualise gradient in real Ghanaian settings.
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- 1Textbook
- 2Exercise book
- 3Ruler and graph board
- 4Calculator
- 5Graph paper
- 6Mini whiteboards
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- 1Learners in pairs compare their four equations written in Phase 2. They check: Is m positive or negative? Is c correct? Partners explain to each other why the gradient tells us if the line is steep or gentle.
- 2Show a new unlabelled graph with a line passing through (1, 4) and (3, 8). Ask: Can you find the gradient? Learners calculate together: (8 − 4) / (3 − 1) = 2. Then they find c by identifying where the line crosses the y-axis visually. They write the complete equation on mini whiteboards and hold them up.
Exercise
- 1The cost of yam at Techiman Market is modelled by the equation y = 2x + 1, where y is cost in Ghana cedis and x is mass in kilograms. (a) Calculate the gradient. (b) State what the gradient represents. (c) Write the y-intercept value. (d) Find the cost when x = 5 kg.
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- 1Recall the linear equation of a graphed line and identify coordinates of points on the line.
- 2Display a completed linear graph showing y = 2x + 3 with five plotted points marked clearly: (0,3), (1,5), (2,7), (3,9), (4,11). Ask learners: What is the y-value when x = 2? Point to the graph and ask them to whisper their answer to a partner. Select one pair to state the coordinate aloud: (2, 7).
- 3Provide learners with five ordered pairs already plotted on a graph: (0, 5), (1, 7), (2, 9), (3, 11), (4, 13). Ask: What pattern do you notice in the y-values? Learners identify that y increases by 2 each time and write this pattern in their exercise books.
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- USING GRAPHS TO FIND MISSING ORDERED PAIRS
- 1Display a completed linear graph for the relation y = 20x (cost of meat in Ghana cedis where x is mass in kilograms and y is total cost). The line passes through (0,0), (1,20), (2,40), (3,60). Point to x = 2.5 on the horizontal axis and ask: What is the y-value? Learners use the ruler and graph board to find where x = 2.5 meets the line vertically and read the y-value (50). They record the ordered pair (2.5, 50) in their exercise books.
- 2Provide learners with a graph showing y = 15x − 10. Using the ruler, learners find the y-values for x = 0.5, x = 1.5, and x = 2.5. They read the coordinates directly from the graph and list them in their exercise books.
- 3Struggling learners: use a grid-aligned graph where x and y values fall exactly on grid intersections only (x = 0, 1, 2, 3). Fast finishers: predict y-values for x = 5 and x = 10 using the equation, then verify by extending their graph line on the graph board.
- 4Emphasise that the graph allows us to read off values without calculating — we just follow up from x and across to the line.
- APPLYING GRAPHS TO REAL PROBLEMS WITH MISSING DATA
- 5Present a real scenario: Abena sells kenkey. The profit she makes is modelled by y = 4x − 5, where y is profit in Ghana cedis and x is number of plates sold. Display the graph for x = 0 to x = 10. Say: Abena sold some plates and made a profit of GH₵15. Using the graph, how many plates did she sell? Learners use the ruler on the graph board to locate y = 15 on the vertical axis, move horizontally to meet the line, then move down to read the x-value (5 plates). They write the ordered pair (5, 15).
- 6Provide learners with three scenarios using the same graph: (i) Abena made GH₵23 profit — how many plates? (ii) She sold 12 plates — what profit? (iii) She made GH₵3 profit — how many plates? Learners use the graph to find the missing x or y value for each. They record all ordered pairs.
- 7Struggling learners: work with the graph and answer guides — they highlight the known value on the axis, then trace to the line and across. Fast finishers: write their own scenario (e.g. distance travelled vs. time) and exchange graphs with a peer to find missing values.
- 8Use the textbook graph examples as models. Remind learners that the graph is their problem-solving tool — they do not need to calculate if they read the graph carefully.
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- 1Textbook
- 2Exercise book
- 3Ruler and graph board
- 4Calculator
- 5Pre-drawn linear graphs
- 6Grid paper
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- 1Show a blank graph with axes labelled x and y. Say: I have the ordered pairs (1, 8), (2, 11), (3, 14). Without drawing, what will be the y-value when x = 4? Learners work out the pattern (gradient = 3) and predict y = 17. One volunteer comes forward and draws the three points and line on the graph board, then locates x = 4 to confirm the y-value is 17.
- 2Learners compare their three scenario answers from Phase 2 with the person next to them. They check: Did we read the graph the same way? Partners explain their method to each other — which axis value did you find first?
Exercise
- 1The graph of y = 3x + 2 is drawn on a grid. Use the graph to find the missing y-value in the ordered pair (4, ___). Then find the missing x-value in the ordered pair (___, 11).
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- 1Analyse a real-life scenario to identify the variables in a linear relationship.
- 2Tell this story: Every morning, Yakubu goes for a walk. After 1 hour, he has walked 3 km. After 2 hours, he has walked 6 km. After 3 hours, he has walked 9 km. Ask learners: What changes as time passes? What stays the same? Learners write two things that change and one that stays the same in their exercise books.
- 3Ask: If we write an equation for Yakubu's walk, which variable represents time and which represents distance? Learners discuss with their partner and one pair volunteers to name them: h = hours (independent), d = distance in km (dependent).
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- SETTING UP LINEAR EQUATIONS FOR REAL-LIFE SITUATIONS
- 1Present the problem: Fatima is a seamstress. She charges GH₵15 for each item sewn, plus GH₵50 for materials. Using the textbook format, set up the equation: Total cost = 15 × (number of items) + 50. Write: C = 15n + 50, where C is cost in cedis and n is number of items. Learners copy the equation and identify m = 15 and c = 50 in their exercise books.
- 2Provide learners with a table showing Fatima's charges: n = 0, C = 50; n = 1, C = 65; n = 2, C = 80; n = 3, C = 95. Using the calculator, they verify each value using the equation C = 15n + 50. They record yes/no for each verification.
- 3Struggling learners: use pre-filled equations with blanks to choose from. Fast finishers: create a second real-life scenario (e.g. trotro fare: GH₵1 per km plus GH₵2 initial charge) and write the equation.
- 4Use familiar Ghanaian contexts — seamstresses, traders, transport — so the equation has meaning.
- USING GRAPHS TO SOLVE REAL PROBLEMS
- 5Draw the graph of C = 15n + 50 on the graph board (x-axis: number of items 0–5; y-axis: cost 0–150). Plot the five points and draw the line. Present the problem: Fatima has a budget of GH₵110. How many items can she sew? Learners use the ruler to locate y = 110 on the vertical axis, find where it meets the line, and read the x-value (4 items). They write the answer in their exercise books: she can sew 4 items for GH₵110.
- 6Provide three more problems using the same graph: (i) How much will 5 items cost? (ii) She has GH₵80. How many items? (iii) She wants to earn enough to charge for 10 items — what will the cost be? Learners read answers directly from the graph for (i) and (ii), and extend the line for (iii).
- 7Struggling learners: work on (i) and (ii) only; the graph grid is aligned to make reading easy. Fast finishers: predict the cost for 12 items without extending the graph — they use the equation C = 15(12) + 50 = 230 and explain why the equation is faster than the graph for large values.
- 8Emphasise: The graph shows us the realistic range. The equation works for any value. This is why real workers use formulas, not just graphs.
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- 1Textbook
- 2Exercise book
- 3Ruler and graph board
- 4Calculator
- 5Graph paper
- 6Mini whiteboards
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- 1Present a new scenario: Kofi delivers water sachets. He earns GH₵0.50 per sachet plus GH₵10 daily allowance. Ask the whole class: What is the equation? Learners write W = 0.5s + 10 on mini whiteboards (where W is earnings and s is sachets sold). Show hands to check agreement.
- 2Ask: If Kofi earned GH₵35 yesterday, how many sachets did he sell? Learners work in pairs to solve using the equation (35 = 0.5s + 10, so s = 50). The first pair to finish brings their work to show. They explain their steps aloud.
Exercise
- 1Ama sells banku at Kaneshie Market. She earns GH₵2 profit per plate plus GH₵8 for transport. (a) Write the linear equation for her total profit P when she sells n plates. (b) Draw the graph of this equation for n = 0 to n = 10 on your graph board. (c) If Ama needs to make GH₵28 profit, how many plates must she sell? Use your graph to find the answer.
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- 1Recall and apply the distributive property with simple single brackets.
- 2Show this physical example: Kwesi has 2 bags with (3 mangoes + 2 oranges) in each. How many pieces of fruit altogether? Learners work it out: 2 × 3 + 2 × 2 = 6 + 4 = 10. Write on the board: 2(3 + 2) = 2 × 3 + 2 × 2 = 10. Ask: What did we do with the 2? We multiplied each item inside the brackets. Learners repeat this sentence aloud three times chorally.
- 3Write on the board: 3(5 + 1). Ask learners to whisper to their partner what the answer is using the pattern they just learned. Partners share: 3 × 5 + 3 × 1 = 15 + 3 = 18. Confirm aloud.
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- EXPANDING SIMPLE BRACKETS USING THE DISTRIBUTIVE PROPERTY
- 1Write on the board: 6(x + 3). Explain: We multiply the 6 by each term inside the brackets. Using the textbook method, write step-by-step: 6 × x + 6 × 3 = 6x + 18. Learners copy this into their exercise books and identify the two parts: 6x (number × variable) and 18 (number × number).
- 2Provide learners with four expressions to expand: (i) 4(x + 2); (ii) 5(y + 4); (iii) 3(2x + 5); (iv) 7(a + 1). Using the calculator to check their multiplications, learners expand each and write the answer. For (iii), emphasise: 3 × 2x = 6x (multiply the numbers, keep the variable).
- 3Struggling learners: work with (i) and (ii) only; provide a box template showing 6(___x + ___3) = ___6x + ___18 to fill in. Fast finishers: expand 8(3x + 2) and then verify by substituting x = 2 into both the original and expanded forms to check they give the same result.
- 4Use the phrase 'distribute the number outside to every term inside' repeatedly. Some learners confuse this with addition — they may write 6x + 3 instead of 6x + 18. Watch for this error.
- EXPANDING BINOMIAL EXPRESSIONS USING DISTRIBUTIVE PROPERTY
- 5Write on the board: (x + 2)(x + 3). Explain: This is the product of two binomials. We use the distributive property twice. First, distribute (x + 3) to x: x(x + 3) = x² + 3x. Then, distribute (x + 3) to 2: 2(x + 3) = 2x + 6. Add them: x² + 3x + 2x + 6 = x² + 5x + 6. Write the working step-by-step on the board. Learners copy into their exercise books carefully.
- 6Provide three binomial expressions: (i) (x + 1)(x + 2); (ii) (x + 3)(x + 4); (iii) (2x + 1)(x + 3). Learners follow the two-step distributive method for each. For (iii), show that they must distribute (x + 3) to 2x first: 2x(x + 3) = 2x² + 6x, then 1(x + 3) = x + 3, then add: 2x² + 6x + x + 3 = 2x² + 7x + 3.
- 7Struggling learners: use a worksheet with the first distribution step already done — they complete the second step and addition. Fast finishers: expand (3x + 2)(2x + 1) and verify their answer by substituting x = 2 into both original and expanded forms.
- 8This is the most complex task of the week. Go slowly. Use coloured pens to show which terms are being distributed. Some learners will need support to understand why we distribute twice.
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- 1Textbook
- 2Exercise book
- 3Calculator
- 4Coloured pens or markers
- 5Worksheets with partially completed expansions
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- 1Learners work in pairs with their exercise books. One learner reads aloud the expression they expanded (e.g. '(x + 2)(x + 3) = x² + 5x + 6'). Their partner checks: Did you distribute to each term? Did you add the middle terms correctly? They switch roles for a second expression.
- 2Show the expression 2(x + 5) on the board. Ask: How many steps do we need to expand this? Learners hold up fingers: 1 step (just one distributive step). Now show (x + 2)(x + 3). Ask: How many steps? Learners hold up fingers: 2 steps (two distributive steps). This reinforces the structure of simple vs. binomial expansion.
Exercise
- 1Expand the following using the distributive property: (a) 5(x + 4); (b) 3(2x + 7); (c) (x + 2)(x + 5). Show all working steps.
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