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The graph of π¦ equals six minus two π₯ is shown for the values of π₯ from zero to three.
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Part a) On the same grid, draw the graph of π¦ equals three minus one-half π₯ for the values of π₯ from zero to six.
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Part b) Using the graphs, solve the simultaneous equations π¦ equals six minus two π₯ and π¦ equals three minus one-half π₯.
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Our first task is to draw π¦ equals three minus one-half π₯ onto the grid.
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To do this, weβll need the coordinates of at least two points.
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However, itβs good to find more than two as that improves the accuracy of our drawing.
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So we make a table for π₯- and π¦-values.
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And we can start by plugging in zero for π₯.
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The equation would say π¦ equals three minus one-half times zero.
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One-half times zero equals zero and three minus zero equals three.
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When π₯ equals zero, π¦ equals three.
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So on our grid, weβll put a point at π₯ equals zero, π¦ equals three, the coordinates zero, three.
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We need at least one other point.
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Iβll choose two.
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I chose to plug in the π₯-coordinate two instead of one because weβre multiplying our π₯-values by one-half.
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And that means if we plug in an even number for our π₯-value, weβll get a whole number for our coordinate.
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So then, we calculate three minus one-half times two.
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One-half times two is one and three minus one is two.
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When π₯ equals two, π¦ equals two.
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On our grid, weβll graph the point two, two.
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Now we could find a third point, where π₯ equals four.
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And so, we solve π¦ equals three minus one-half times four.
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One-half times four equals two and three minus two equals one.
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When π₯ equals four, π¦ equals one.
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Now weβll graph our point four, one.
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If we look closely, weβll notice that weβre going down one along the π¦-axis and right two along the π¦-axis.
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Using this method, we can find our final point down one and right two.
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We put a point here and we see that when π₯ equals six, π¦ equals zero.
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Our question was asking us to draw this graph from zero to six.
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To do that, weβll sketch a line across these four points.
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Itβs worth noting here that we could put our functions into π¦-intercept form: π¦ equals ππ₯ plus π.
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The first function we were given that was already graphed is π¦ equals six minus two π₯.
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If we rearrange it to π¦ equals ππ₯ plus π, it would be π¦ equals negative two π₯ plus six.
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If you wanted to graph using slope-intercept form, you start at the π¦-intercept.
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Here thatβs six.
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Since we have a slope of negative two, we can think of this as negative two over one.
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This tells us we need to go down two and right one.
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Our second function π¦ equals three minus one-half π₯ can be rewritten as π¦ equals negative one-half π₯ plus three.
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To graph this function, start at the π¦-intercept of three and then we go down one and right two because the slope is negative one-half.
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And we know that the slope or the gradient is the changes in π¦ over the changes in π₯.
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The changes in π¦ of this function is negative one and the changes in π₯ is positive two β down one, right two.
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Moving on to part b), using the graph, we want to solve the simultaneous equations.
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Solving simultaneous equations is finding the coordinate that falls on both lines, the point where they intersect.
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When we look at our graph, we see that both of these lines have a point at two, two.
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When π₯ equals two, π¦ equals two for both of these lines.
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However, itβs probably good to go ahead and check that this is true.
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For our first function π¦ equals six minus two π₯, we plug in two for our π₯- and π¦-values.
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So we have two equals six minus two times two.
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Two times two is four.
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Six minus four is two.
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Two equals two.
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And that means this statement is true for the first graph.
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We plug in two for our π₯ and π¦.
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And we have two equals three minus one-half times two.
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One-half times two equals one.
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Three minus one equals two.
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And so, the point two, two also lies on our second graph.
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Using the graph, we found that the simultaneous equations have a solution at two, two.