WEBVTT
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Which of the following sets of simultaneous equations could be solved using the given graph?
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(a) π¦ equals two π₯ minus four and π¦ equals π₯ plus five.
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(b) π¦ equals negative four π₯ plus two and π¦ equals five π₯ minus one.
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(c) π¦ equals two π₯ minus four and π¦ equals negative π₯ plus five.
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(d) π¦ equals two π₯ plus four and π¦ equals negative π₯ plus five.
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Or (e) π¦ equals negative four π₯ plus two and π¦ equals five π₯ plus one.
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So, weβve been given a graph of two straight lines and asked to determine which pair of simultaneous equations we could solve using this graph.
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This means that we need to determine the equations of the two straight lines.
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In order to do this, weβll recall the general form of a straight line in its slopeβintercept form π¦ equals ππ₯ plus π.
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And we recall that the coefficient of π₯, thatβs π, gives the slope of the line.
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And the constant term, thatβs π, gives the π¦-intercept of the graph.
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Thatβs the π¦-value at which the line intercepts the π¦-axis.
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Letβs consider the blue line first of all.
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We can see that this line intercepts the π¦-axis at five, which means the equation of this line will be in the form π¦ equals some number of π₯ plus five.
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To find the slope of this line, the value of π, we can draw in a little right-angled triangle anywhere below this line.
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And in doing so, we see that for every one unit the line moves across, thatβs to the right, it also moves one unit down.
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As the slope of a line can be found using change in π¦ over change in π₯, we have negative one over one, which is negative one.
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The equation of this line is, therefore, π¦ equals negative π₯ plus five.
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So, we found the equation of our first line.
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Letβs now consider the red line.
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And this time we see that this line intercepts the π¦-axis at a value of negative four.
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The equation of this line is, therefore, in the form π¦ equals some number of π₯ minus four.
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To find the slope, we again sketch in a right-angled triangle anywhere below this line.
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And this time, we see that for every one unit the line moves to the right, it moves two units up.
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The slope of the line, change in π¦ over change in π₯, is therefore two over one, which is equal to two.
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So, the equation of this line is π¦ equals two π₯ minus four.
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Looking at the five possible options we were given, we can see that this combination of equations of straight lines is option (c) π¦ equals two π₯ minus four and π¦ equals negative π₯ plus five.
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Although we arenβt actually asked to solve the pair of simultaneous equations in this question, we could do so by looking at the coordinates of the point of intersection of the two lines.
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And we see that the coordinates of this point are three, two.
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So, the solution to this pair of simultaneous equations would be π₯ equals three and π¦ equals two.