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What is the value of π for which the simultaneous equations π₯ plus two π¦ equals one and five π₯ plus ππ¦ equals five do not have a unique solution?
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Equations that do not have a unique solution do not intersect, which means they are parallel lines and they have equals slopes.
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To find the slope, letβs take both of these equations and put them in the slope-intercept form.
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Thatβs the form π¦ equals ππ₯ plus π.
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And in this form, π represents the slope.
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To get equation one in the slope-intercept form, we need to isolate π¦.
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We can do that by subtracting π₯ from both sides of the equation.
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We then have two π¦ equals one minus π₯.
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But because weβre trying to get the form π¦ equals ππ₯ plus π, we can flip the π₯ and the π.
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It will then say two π¦ equals negative π₯ plus one.
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From there, we divide everything by two.
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π¦ equals negative π₯ over two plus one-half.
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But remember weβre looking for that π value.
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So we wanted to say π¦ equals negative one-half times π₯ plus one-half.
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π equals the slope and it equals negative one-half.
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We want to get our second equation in the same form.
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So we subtract five π₯ from both sides.
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ππ¦ equals five minus five π₯.
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We want our π₯ term first.
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So weβll flip them around, being careful to keep that negative with our π₯.
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ππ¦ equals negative five π₯ plus five.
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After that, we want π¦ by itself.
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So we divide everything by π.
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π¦ equals negative five over ππ₯ plus five over π.
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The slope of our second equation is negative five over π.
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And we want to know what π is.
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These two lines do not have a unique solution.
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And that means they have equal slopes.
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Negative one-half equals negative five over π.
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Letβs get the π out of the denominator, multiply by π on both sides.
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π times negative one equals negative π.
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The denominator stays negative two.
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And that equals negative five.
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To get two out of the denominator, we multiply by two on both sides.
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Negative π equals negative 10.
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And that means positive π equals 10.
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When π equals 10, these two lines have the same slope.
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And that means they have no unique solution.
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But before we leave this problem, I want to show you one other way to think about it.
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This time, weβre going to look at the coefficients of π₯ and π¦ in both of these equations.
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And weβre going to make a ratio of these coefficients.
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The coefficient of π₯ in the first equation is one.
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We could say one π₯.
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And the coefficient of π¦ is two.
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They have a ratio of one to two.
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In the second equation, the coefficient of π₯ is five and the coefficient of π¦ is π.
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Notice that to get from one to two, we multiply by two.
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And if we want these two ratios to be equal, then to go from five to π, we will also multiply by two.
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Five times two is 10.
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So in order for these equations to be parallel, π must be equal to 10.
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Both methods are equally valid.
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One just requires a little bit more algebra than the other.