WEBVTT
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Using elimination, solve the simultaneous equations three π₯ plus seven π¦ equals 34 and nine π₯ plus 10 π¦ equals 91.
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Our first step is to make either the π₯ or the π¦ coefficients the same.
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In this case, the easiest way to do this is to multiply the first equation by three.
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Multiplying three π₯ by three gives us nine π₯.
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Multiplying seven π¦ by three gives us 21 π¦.
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And 34 multiplied by three is 102.
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If we then subtract equation two from equation one, the π₯ terms cancel as nine π₯ minus nine π₯ is zero.
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21 π¦ minus 10 π¦ is equal to 11 π¦.
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And 102 minus 91 is equal to 11.
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Dividing both sides of this equation by 11 gives us an answer for π¦ equal to one.
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In order to work out our value for π₯, we need to substitute this value for π¦ into one of the equations.
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In this case, Iβm going to substitute π¦ equals one into equation two.
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Substituting in this value for π¦ gives us nine π₯ plus 10 multiplied by one equals 91.
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As 10 multiplied by one is 10, weβre left with nine π₯ plus 10 equals 91.
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We can then subtract 10 from both sides of the equation, leaving us nine π₯ is equal to 81.
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And finally dividing both sides of this equation by nine leaves as a value for π₯ equal to nine.
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Therefore, the solution to the simultaneous equations three π₯ plus seven π¦ equals 34 and nine π₯ plus 10 π¦ equals 91 are π¦ equals one and π₯ equals nine.
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We could check these two answers by substituting the values of π₯ and π¦ back into the other equation, number one.
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Nine multiplied by nine plus 21 multiplied by one is equal to 102.
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Therefore, our solution is correct.