WEBVTT
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This question has three parts.
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Part a) Complete the table of values for π¦ equals four minus two π₯.
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Part b) Draw the graph of π¦ equals four minus two π₯ on the grid for the values of π₯ from negative two to four.
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And c) Solve π₯ equals four minus two π₯.
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Starting with part a), we know that our π¦-values equal four minus two times our π₯-values.
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If we want to find π¦ when π₯ equals negative two, we substitute π₯ in our equation with the value negative two.
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By order of operations, we should multiply negative two times negative two first, which is positive four.
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In this case, π¦ equals four plus four.
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π¦ equals eight.
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Next, we want to consider what π¦ is if π₯ is zero.
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So we plug in zero for our π₯-value.
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Two times zero equals zero.
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And that means when π₯ equals zero, π¦ equals four.
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And we plugged that into our table.
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We could continue using the substitution method.
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However, at this point, we also notice something else.
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Every time π₯ is increased by one, π¦ is decreased by two.
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And that means we can find our next value on the table by taking two and subtracting two.
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When π₯ equals two, π¦ equals zero.
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We can confirm this with substitution.
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Two times two equals four.
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And four minus four equals zero.
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Again, we can just subtract two.
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Zero minus two is negative two.
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And negative two minus two equals negative four.
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If youβre going to use this method, you need to check and make sure that your π₯-values are increasing by the same amount each time.
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Now that weβve completed our table, part b) wants us to draw the graph of π¦ equals four minus two π₯.
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We can use some of the points for our table to help us.
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We can use a few of the coordinates from our table to help us sketch this graph.
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When π₯ equals negative two, π¦ equals eight.
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When π₯ equals zero, π¦ equals four.
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So we have a point at zero, four.
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When π₯ equals two, π¦ equals zero.
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And we have a point at two, zero.
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And finally, when π₯ equals four, π¦ equals negative four.
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Weβll sketch a line to connect these four points.
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This is the graph of π¦ equals four minus two π₯, from negative two to four.
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Part c) is asking us to solve π₯ equals four minus two π₯.
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We wanna solve for π₯.
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But we have an π₯-variable on both sides of the equation.
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And we need to try and get them on the same side.
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To do that, weβll add two π₯ to both sides of our equation.
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π₯ plus two π₯ equals three π₯.
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And negative two π₯ plus two π₯ cancels out.
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So weβre left with four on the right side.
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We can now see that three π₯ equals four.
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We then divide by three on both sides of the equation.
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Four divided by three we can write as four over three, four-thirds.