To graph linear inequalities, first identify the boundary line type: use solid lines for 'greater than or equal to' or 'less than or equal to' inequalities, and dotted lines for strict inequalities. For vertical lines (x = constant), draw a line at that x-value. For lines in slope-intercept form (y = mx + c), plot the y-intercept and use the gradient to find additional points. When equations are not in slope-intercept form, rearrange them by isolating y on one side. The solution region is the area that satisfies all given inequalities simultaneously.
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Prérequis
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Prochaines étapes
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Graphical Inequalities Part 1/2 | GCSE Maths 2026Ajouté :
Shading regions with graphical inequalities, tricky topic, not many people like it. Let's see what we can do. This one is four marks. On the grid show by shading the region that satisfies all of these inequalities.
Label the region R. The first thing I recognize is that they're all greater than or equal to or less than or equal to. So, they're all going to be solid lines. If it was less than or greater than, we'd have the dotted line.
But, because we have the or equal to, we have a solid line. So, the first one to draw is x = 0. And we're going to draw each of them as um having equal signs first and then decide whether we want above or below the line.
So, first off, x = 0.
This is a vertical line through the uh y-axis because it's where x is always equal to 0. So, that one's on there.
x is less than or equal to 2. So, this time a vertical line through 2 because this is where x is always equal to 2.
Next up, um y = x + 3. y-intercept is 3, gradient is 1. So, we can draw on the y-intercept of 3, then we go one across, one up, one across, one up, and we can draw that line on there joining those points.
So, this is y = x + 3.
This is x = 0. And this is x = 2. And now for sketching this last one, it's going to be a little bit annoying. Why?
Because we don't have this in the form y = mx + c to be easily sketched. So, what we're going to do is we're going to firstly rearrange this to the form y = mx + c.
So, we have 2x. Actually, I'll do it over to the left-hand side.
We have 2x + 3y = 6.
3y = -2x + 6. Divide everything through by 3. y = - 2/3 x + 2.
Any questions for that one, guys? Let me know.
Now, what can we do? We can sketch this on. Y intercept is 2. So, we
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