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Use the graph to identify which of the given points is not a solution to the given set of inequalities: 𝑦 is less than two 𝑥 plus four, 𝑦 is greater than or equal to negative three 𝑥, 𝑥 is less than or equal to two.
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Our options are A) the point two, negative six; B) the point zero, zero; C) the point one, six; D) the point two, negative two; and E) the point two, two.
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We have a set of three inequalities, and a solution to the set of inequalities must satisfy each inequality individually.
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The point that we’re looking for therefore is the odd one out; it’s the point which doesn’t satisfy one or more of these inequalities and so doesn’t satisfy the given set of inequalities.
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We could substitute the values of 𝑥 and 𝑦 for each point into each inequality to see if they satisfy them, but we’re told in the question that we should use the graph, so let’s do that.
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What does it mean to represent the first inequality, 𝑦 is less than two 𝑥 plus four on the graph?
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Well the set of points which satisfy 𝑦 equals two 𝑥 plus four form a line on this graph.
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We’re looking for the points for which 𝑦 is less than two 𝑥 plus four, and so the points must be below this line.
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And you can check that a point below this line really does satisfy 𝑦 is less than two 𝑥 plus four and a point above this line will satisfy 𝑦 is greater than two 𝑥 plus four.
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And you might be able to notice that the region below this line is darker than the region above this line, and that’s because the region below the line has been shaded.
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To satisfy the set of inequalities, you must also satisfy the second inequality, 𝑦 is greater than or equal to negative three 𝑥.
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We can draw the line 𝑦 equals negative three 𝑥.
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Any point on this line will satisfy the equation 𝑦 equals negative three 𝑥, and any point below this line will satisfy the inequality 𝑦 is less than negative three 𝑥.
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And so the inequality 𝑦 is greater than or equal to negative three 𝑥 is satisfied by any point which is either on the line or above the line.
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And finally, we have the inequality 𝑥 is less than or equal to two.
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Here is the line 𝑥 equals two.
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Any point on this line or to the left of this line will satisfy 𝑥 is less than or equal to two.
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The feasible region is the region for which all the inequalities are satisfied, so that means we have to be below the orange line, above the purple line, and to the left of the green line.
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We can shade this region blue now.
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Remember that we’re looking for the point which is not a solution to the set of inequalities, so this is a point which is not in the region that we’ve shaded.
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Let’s have a look at the point two and negative six.
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This is on one of the vertices of the shaded region.
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The question is is it inside the region or not.
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As the two lines that it lies on are both solid and not dashed or dotted, we say that it is.
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The fact that these two lines are solid tells us that the inequalities they correspond to are inclusive, so we had 𝑦 is greater than or equal to negative three 𝑥 and 𝑥 is less than or equal to two.
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As each inequality has that “equal to” bit, the points on those lines will satisfy the inequalities, and we draw those lines using a solid line to remind ourselves of that fact.
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So A is in our feasible region and so isn’t our answer.
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We move onto B.
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B is the point zero, zero, which we can mark on our diagram.
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Again this is on the boundary of our feasible region, so we have to think slightly carefully about whether this point should be included or not.
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But as it lies on a solid line and not a dashed line, it is included.
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This point is clearly below the line 𝑦 equals two 𝑥 plus four, so it must satisfy the inequality 𝑦 is less than two 𝑥 plus four.
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Similarly it’s very clearly to left of the line 𝑥 equals two, so it must satisfy the inequality 𝑥 is less than or equal to two.
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The only question was about the inequality 𝑦 is greater than or equal to negative three 𝑥.
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As it lies on the line 𝑦 is equal to negative three 𝑥, and so it does satisfy all three inequalities.
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We move onto the point one, six, which we can plot on our graph.
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This point again lies on the boundary of our feasible region, and so the question is does it satisfy all three inequalities.
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Notice that this point lies on the line 𝑦 equals two 𝑥 plus four, which is dashed.
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This line is dashed because it represents the strict inequality 𝑦 is less than two 𝑥 plus four.
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As this inequality is strict, points on this line do not satisfy the inequality, and we dashed the line to show this.
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And so as this point lies on this dashed line, it doesn’t satisfy the corresponding inequality.
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Our answer is therefore that the point one, six is not a solution to the given set of inequalities because it doesn’t satisfy the first of the inequalities, 𝑦 is less than two 𝑥 plus four.
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And we can just check that the other two points D, which is the point two and negative two, and E, which is point two, two, do satisfy all three inequalities.
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You can see that they lie on the line 𝑥 equals two, and this line is solid and so is part of the shaded region.