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Solve the following equation: ๐ฅ over three plus ๐ฅ minus 14 over two ๐ฅ minus five is equal to two.
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Give your solutions to two decimal places.
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You must show your working.
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Letโs remind ourselves how we add fractions.
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Usually, we find the lowest common denominator of the two fractions.
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We donโt know the lowest common denominator for these two fractions since theyโre algebraic.
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But we can find a common denominator.
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To find a common denominator, we find the product of the two denominators.
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In this case, a common denominator is three multiplied by two ๐ฅ minus five.
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To expand these brackets, we need to multiply each term on the inside by the three on the outside.
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Three multiplied by two is six.
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So three multiplied by two ๐ฅ is six ๐ฅ.
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Three multiplied by five is 15.
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So three multiplied by negative five is negative 15.
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And our common denominator is six ๐ฅ minus 15.
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Remember though when we create the common denominator, we need to ensure that weโre also multiplying the numerator to form equivalent fractions.
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We multiply the denominator of the first fraction by two ๐ฅ minus five.
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So we need to do that to the numerator.
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That gives us ๐ฅ multiplied by two ๐ฅ minus five all over six ๐ฅ minus 15.
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Once again, we can expand by multiplying each term inside this bracket by the ๐ฅ on the outside.
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๐ฅ multiplied by two ๐ฅ is two ๐ฅ squared and ๐ฅ multiplied by negative five is negative five ๐ฅ.
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Similarly, we multiply the denominator of the second fraction by three.
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So we also need to multiply the numerator by three.
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Thatโs three multiplied by ๐ฅ minus 14 all over six ๐ฅ minus 15.
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Once again, we expand the brackets on the numerator as before.
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And we get three ๐ฅ minus 42 all over six ๐ฅ minus 15.
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Now that we have two fractions with the same denominator, we can add their numerators.
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We have two ๐ฅ squared minus five ๐ฅ over six ๐ฅ minus 15 plus three ๐ฅ minus 42 over six ๐ฅ minus 15.
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Thatโs two ๐ฅ squared minus five ๐ฅ plus three ๐ฅ minus 42 all over six ๐ฅ minus 15.
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And of course, we can simplify the numerator by collecting like terms.
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Negative five ๐ฅ plus three ๐ฅ is negative two ๐ฅ.
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And the expression on the left-hand side of our original equation becomes two ๐ฅ squared minus two ๐ฅ minus 42 all over six ๐ฅ minus 15.
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Letโs put this back into the original equation.
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Weโre going to need to clear some space.
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We have two ๐ฅ squared minus two ๐ฅ minus 42 over six ๐ฅ minus 15 is equal to two.
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Now, this doesnโt look particularly nice.
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But we can get rid of the nasty looking denominator on the left-hand side by multiplying through by six ๐ฅ minus 15.
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Remember if we do that, we cancel out the denominator on the left-hand side.
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We donโt then need to multiply the numerator by six ๐ฅ minus 15.
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So weโre just left with two ๐ฅ squared minus two ๐ฅ minus 42.
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And thatโs equal to two multiplied by six ๐ฅ minus 15.
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Expanding those brackets like we have been, we get 12๐ฅ minus 30 on the right-hand side of this equation.
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Now, you might have noticed that we have a quadratic expression on the left-hand side.
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We canโt solve this until we make sure that we have a quadratic equation thatโs equal to zero.
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So letโs begin by subtracting 12๐ฅ from both sides.
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That gives us two ๐ฅ squared minus 14๐ฅ minus 42 is equal to negative 30.
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Weโre then gonna add 30 to both sides of the equation.
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And we get two ๐ฅ squared minus 14๐ฅ minus 12 is equal to zero.
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Notice how each term in this equation is a multiple of two.
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So we can divide everything by two.
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And if we do that, we get ๐ฅ squared minus seven ๐ฅ minus six is equal to zero.
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Now that our quadratic equation is in the right form, we can solve it.
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We usually try to factorise.
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But there is a hint that this might not be suitable in the question.
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Weโre told to give our solutions to two decimal places.
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So that tells us that the expression ๐ฅ squared minus seven ๐ฅ minus six will not be factorable.
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Instead, weโll need to use the quadratic formula and this is something you need to know by heart.
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For a quadratic equation of the form ๐๐ฅ squared plus ๐๐ฅ plus ๐ is equal to zero, ๐ฅ has two solutions.
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Itโs negative ๐ plus or minus the square root of ๐ squared minus four ๐๐ all over two ๐.
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Letโs compare our equation to the general form ๐๐ฅ squared plus ๐๐ฅ plus ๐.
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๐ is the coefficient of ๐ฅ squared.
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Itโs the number of ๐ฅ squareds we have, which in this case is one.
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๐ is the coefficient of ๐ฅ.
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Itโs the number of ๐ฅs we have.
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In this case, thatโs negative seven.
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And ๐ is the constant.
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Here, thatโs negative six.
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Letโs substitute these into the formula for the quadratic equation.
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Doing so and we get negative negative seven plus or minus the square root of negative seven squared minus four multiplied by one multiplied by negative six all over two multiplied by one.
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Youโll notice that all the negative numbers are in brackets here.
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Thatโs particularly important for the negative seven squared.
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It ensures your calculator knows itโs squaring a negative number.
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We can split it up and type in the addition and subtraction part as shown.
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But it is sensible to evaluate each part first.
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Negative negative seven is seven, negative seven squared is 49, and negative four multiplied by one multiplied by negative six is 24.
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So we have seven plus or minus the square root of 49 plus 24 over two.
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Letโs calculate the addition part first.
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If we type this into our calculator, we get 7.7720.
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And then, the second solution for ๐ฅ is the subtraction part.
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And that gives us negative 0.7720.
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Remember we need to give our answers to two decimal places.
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In each number, the second digit after the decimal point is a seven.
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We look to the digit immediately to the right of this.
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This is called the deciding digit.
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Since in both cases, the deciding digit is two, itโs less than five, we round the number down.
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And weโve solved the equation.
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Correct to two decimal places, ๐ฅ is equal to 0.77 or negative 0.77.