WEBVTT
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Find the equation of the straight line that passes through the point of intersection of the two straight lines π₯ minus eight π¦ equals two and negative six π₯ minus eight π¦ equals one and is parallel to the π¦-axis.
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So weβre looking for the equation of this particular straight line.
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And the first thing we notice that this line is parallel to the π¦-axis.
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Now we can recall that lines which are parallel to the π¦-axis are vertical lines.
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And they therefore have equations of the form π₯ equals π, where π is some constant.
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And π is the value at which these lines cross the π₯-axis.
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We also know that the straight line weβre looking for passes through the point of intersection of the two straight lines whose equations weβve been given.
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So what weβre going to need to do is solve these two equations simultaneously in order to find their point of intersection.
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Or at least find the π₯-coordinate of their point of intersection.
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So these are the two equations weβre looking to solve.
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Theyβre linear simultaneous equations in two variables.
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There are a couple of different approaches we could take.
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Firstly, we could note that both equations have the same coefficient of π¦.
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They both have negative eight π¦.
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And so we could use the acronym SSS, standing for same sign subtract, in order to eliminate the π¦-variables.
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If we subtract equation two from equation one, then we have π₯ minus negative six π₯, which is π₯ plus six π₯ or seven π₯.
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We have negative eight π¦ minus negative eight π¦.
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Thatβs negative eight π¦ plus eight π¦.
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So the π¦s are eliminated.
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And on the right-hand side, we have two minus one, which is equal to one.
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We therefore have the equation seven π₯ equals one, and weβve eliminated the π¦-variable.
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To solve this equation, we just need to divide both sides by seven, giving π₯ equals one-seventh.
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An alternative method would be to use the method of substitution.
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We could rearrange equation one to give negative eight π¦ is equal to two minus π₯.
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And we can then substitute this expression for negative eight π¦ into equation two.
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Doing so gives negative six π₯ plus two minus π₯ is equal to one.
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And now we have an equation in π₯ only.
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We can then simplify to give negative seven π₯ plus two equals one.
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Subtract two from each side to give negative seven π₯ equals negative one, and then divide by negative seven.
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Giving π₯ equals negative one over negative seven, which is equal to one-seventh.
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In both cases then, we found that the π₯-coordinate of the point of intersection of these two straight lines is one-seventh.
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Now we donβt need to go any further because we donβt need to know the π¦-coordinate of the point of intersection.
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Remember, we said straight lines parallel to the π¦-axis have equations of the form π₯ equals some constant.
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So if our line passes through the point where π₯ equals one-seventh, its equation must just be π₯ equals one-seventh.
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So by first recalling the general equation of a straight line which is parallel to the π¦-axis and then partially solving the equations of the two straight lines to find the π₯-coordinate of their point of intersection.
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Weβve found that the equation of the straight line that passes through their point of intersection and is parallel to the π¦-axis is π₯ equals one-seventh.