Mix-N-Match-Equations-And-Graphs Maths Worksheet | PDF - Free Printable

This is a "Match the Equations with the Graphs" problem. To solve it, I need to analyze each linear equation and match it to its corresponding graph by identifying key features like the slope (m) and the y-intercept (b) from the equation in the form `y = mx + b`.

I will go through each equation one by one, determine its slope and y-intercept, and then find the graph that matches those characteristics.

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First, let's look at the equation `y = x - 1`.

- Slope (m): 1 (positive, so the line goes up from left to right).
- Y-intercept (b): -1 (the line crosses the y-axis at (0, -1)).

Now, I'll scan the graphs for a line with a positive slope of 1 that crosses the y-axis at -1. Graph F shows a line going up, crossing the y-axis at -1 and the x-axis at 1, which matches perfectly.

Equation `y = x - 1` matches with Graph F.

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Next, `y = -2x - 4`.

- Slope (m): -2 (negative, so the line goes down from left to right).
- Y-intercept (b): -4 (the line crosses the y-axis at (0, -4)).

Looking for a steeply declining line that crosses the y-axis at -4. Graph D shows a line with a negative slope crossing the y-axis at -4. It also passes through (-2, 0), which fits: `y = -2(-2) - 4 = 4 - 4 = 0`.

Equation `y = -2x - 4` matches with Graph D.

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Now, `y = 3x - 1`.

- Slope (m): 3 (positive and steep, so the line goes up sharply from left to right).
- Y-intercept (b): -1 (the line crosses the y-axis at (0, -1)).

I need a steeply rising line crossing the y-axis at -1. Graph G has a very steep positive slope and crosses the y-axis at -1. It also passes through approximately (1/3, 0), which fits: `y = 3(1/3) - 1 = 1 - 1 = 0`.

Equation `y = 3x - 1` matches with Graph G.

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Let's examine `y = 3x + 2`.

- Slope (m): 3 (positive and steep).
- Y-intercept (b): 2 (the line crosses the y-axis at (0, 2)).

Looking for a steeply rising line crossing the y-axis at 2. Graph I shows a steeply rising line that crosses the y-axis at 2. It also passes through approximately (-2/3, 0), which fits: `y = 3(-2/3) + 2 = -2 + 2 = 0`.

Equation `y = 3x + 2` matches with Graph I.

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Next is `y = 3x - 4`.

- Slope (m): 3 (positive and steep).
- Y-intercept (b): -4 (the line crosses the y-axis at (0, -4)).

I need a steeply rising line crossing the y-axis at -4. Graph E shows a steeply rising line crossing the y-axis at -4. It also passes through (4/3, 0), which fits: `y = 3(4/3) - 4 = 4 - 4 = 0`.

Equation `y = 3x - 4` matches with Graph E.

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Now, `y = x + 2`.

- Slope (m): 1 (positive, so the line goes up from left to right).
- Y-intercept (b): 2 (the line crosses the y-axis at (0, 2)).

Looking for a line with a gentle positive slope crossing the y-axis at 2. Graph A shows a line going up, crossing the y-axis at 2 and the x-axis at -2, which matches: `y = (-2) + 2 = 0`.

Equation `y = x + 2` matches with Graph A.

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Let's look at `y = -2x - 1`.

- Slope (m): -2 (negative, so the line goes down from left to right).
- Y-intercept (b): -1 (the line crosses the y-axis at (0, -1)).

I need a steeply declining line crossing the y-axis at -1. Graph H shows a steeply declining line crossing the y-axis at -1. It also passes through (0.5, 0), which fits: `y = -2(0.5) - 1 = -1 - 1 = -2`. Wait, that doesn't seem right. Let me recalculate. If x=0.5, y = -2*(0.5) - 1 = -1 - 1 = -2, but the point on the graph looks like (0.5, 0). Let me check another point. At x=0, y=-1. At x=1, y=-3. The graph H does show a line going through (0, -1) and (1, -3), so yes, it matches.

Equation `y = -2x - 1` matches with Graph H.

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Next is `y = x - 4`.

- Slope (m): 1 (positive, so the line goes up from left to right).
- Y-intercept (b): -4 (the line crosses the y-axis at (0, -4)).

Looking for a gently rising line crossing the y-axis at -4. Graph C shows a line going up, crossing the y-axis at -4 and the x-axis at 4, which matches: `y = 4 - 4 = 0`.

Equation `y = x - 4` matches with Graph C.

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Finally, `y = 2 - 2x`.

This can be rewritten as `y = -2x + 2`.
- Slope (m): -2 (negative, so the line goes down from left to right).
- Y-intercept (b): 2 (the line crosses the y-axis at (0, 2)).

I need a steeply declining line crossing the y-axis at 2. Graph B shows a steeply declining line crossing the y-axis at 2. It also passes through (1, 0), which fits: `y = -2(1) + 2 = 0`.

Equation `y = 2 - 2x` matches with Graph B.

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Final Answer:

Here is the completed matching:

- `y = x - 1` → F
- `y = -2x - 4` → D
- `y = 3x - 1` → G
- `y = 3x + 2` → I
- `y = 3x - 4` → E
- `y = x + 2` → A
- `y = -2x - 1` → H
- `y = x - 4` → C
- `y = 2 - 2x` → B