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Maths Summer School

9 Algebra exercises – Solutions

This section contains solutions to the bonus exercises to accompany the Algebra chapter. Each chapter contained embedded examples and exercises, often with explanations.

These solutions are provided with fewer explanations, generally just with the basic answers provided later. Of course, you can ask questions of the course lecturer about any problems here too.

9.1 Expansion of brackets and powers

  1. Expand the following expressions involving products and powers
    1. \(\left(3x^2\right)\left(5x^4\right)=15x^6\)
    2. \(\left(8x^3 \right)\left(9x^5 \right)=72x^8\)
    3. \(\left( 4xy \right)\left( 7x^2y^3 \right)=28x^3y^4\)
    4. \(\left( -x^3y^4 \right)\left(6x^8y^9 \right)=-18x^{11}y^{13}\)
    5. \(\left(5x^2y^7 \right)\left(-7x^3y^9 \right)=-35x^5y^{16}\)
    6. \(\left(3ab \right)^2\left(-5ab^3 \right)=-45a^3b^3\)
    7. \((-5x^2y)(-8x^3y^2)=40x^5y^3\)
    8. \(\left(-3a^b \right)\left(-4ab^3 \right)=12a^3b^4\)
    9. \(\left(4a \right)\left(ab \right)\left(-a^2b\right)=-4a^4b^2\)
    10. \(\left(-3ab \right)\left(-4a^2b \right)\left(ab^2 \right)^2=12a^5b^6\)
  2. Expand the following brackets and simplify
    1. \((x+2)(x+3)=x^2+5x+6\)
    2. \((2x+5)(3x-7)=6x^2+x-35\)
    3. \((2x-3)(2x+3)=4x^2-9\)
    4. \((4y-8)(6y-13)=24y^2-100y+104\)
    5. \((x+4)^2=x^2+8x+16\)
    6. \((2x+3y)^2=4x^2+12xy+9y^2\)
    7. \((2x-3y)^2=4x^2-12xy+9y^2\)
    8. \((x+2)(x^2-4x+5)=x^3-2x^2-3x+10\)
    9. \((2x-1)(x^2-3x+7)=2x^3-7x^2+17x-7\)
    10. \((x+3)(x-4)(x+5)=x^3+4x^2-17x-60\)
  3. Expand the brackets \((a-b)(a+b)\) and then use this general formula to answer the following without doing any new algebra \((a-b)(a+b)=a^2-b^2\) so…
    1. \((x+2)(x-2)=x^2-4\)
    2. \((2x-3)(2x+3)=4x^2-9\)
    3. \(x^2-16=(x-4)(x+4)\)
    4. \(25x^2-64=(5x-8)(5x+8)\)

9.2 Simplifying fractions

  1. Simplify the following fractions containing powers
    1. \[\frac{12a^2}{3a^2}=4\]
    2. \[\frac{-32b^3}{8b^2}=-4b\]
    3. \[\frac{42a^3b^4}{6a^2b}=7ab^3\]
    4. \[\frac{72c^2d^3}{-9cd^2}=-8cd\]
    5. \[\frac{4xy-8xy^2}{2xy}=2-4y\]
    6. \[\frac{6\pi rh^2+18\pi r^2h}{3\pi r h}=2h+6r\]

9.3 Exponentials and Logarithms

  1. Use the \(\log\) button your calculator (which is actually base \(10\), i.e. \(\log_{10}\)) and the \(10^{\Box}\) button to evaluate the following:

    1. \(\log_{10}(1000)=3\)
    2. \(\log_{10}(2.5)=0.397940009\)
    3. \(10^{3.5}=3162.27766\)
    4. \(\log_{10}(10^{3.5})=3.5\)
    5. \(\log_{10}(10^{-1.4})=-1.4\)
    6. \(10^{\log_{10}(6.1)}=6.1\)
    7. \(10^{\log_{10}(0.5)}=0.5\)

    Could you have known the answer to any of these without using your calculator?

  2. Use the \(\ln\) (which is just \(\log_e\)) on your calculator, and its counterpart \(e^{\Box}\) to evaluate the following:

    1. \(\ln(4.5) =1.504077397\)
    2. \(\ln(2.75)=1.011600912\)
    3. \(e^{0.6}=1.822118800\)
    4. \(e^{-1.5}=0.223130160\)
    5. \(\ln(e^{0.6})=0.6\)
    6. \(\ln(e^{-1.5})=-1.5\)
    7. \(e^{\ln(2.5)}=2.5\)
    8. \(e^{\ln(0.75)}=0.75\)

    Could you have known the answer to any of these without using your calculator?

  3. Use the three log laws to expand these logarithms into sums and differences of simpler logarithms.

    1. \(\log_{10}\left(\frac{3x^2}{y} \right)= \log_{10}(3)+2\log_{10}(x)-\log_{10}(y)\)
    2. \(\ln\left( \frac{x^2y^2}{4} \right) =2\ln(x)+2\ln(y)-\ln(4)\)
    3. \(\log_{10}\left( \frac{100}{x+1}\right) =2-\log_{10}(x+1)\)
    4. \(\ln\left( \frac{e^2}{2x+3} \right) =2-\ln(2x+3)\)
    5. \(\log_{10}\left( \sqrt{x^2+1} \right) =\frac{1}{2}\log_{10}(x^2+1)\)
    6. \(\ln\left(\sqrt{\frac{(x+1)^3(x-1)}{(x+2)}} \right) =\frac{1}{2}\left[ 3\ln(x+1)+\ln(x-1)-\ln(x+2) \right]\)

  4. Combine these sums and differences of simple logs into one single logarithm:

    1. \(3\log_{10}(x)+2\log_{10}(y)-4\log_{10}(z)=\log_{10}\left(\frac{x^3y^2}{z^4}\right)\)
    2. \(2\log_{10}(x+y)-\frac{1}{2}\log_{10}(z)=\log_{10}\left(\frac{\left(x+y\right)^2}{\sqrt{z}}\right)\)
    3. \(3\ln(x)+\frac{1}{3}\ln(y)=\ln\left(x^3y^{\frac{1}{3}}\right)\)
    4. \(4\ln(2x+y)-2\ln(z)=\ln\left(\frac{\left(2x+y\right)^4}{z^2} \right)\)

9.4 Re-arranging formula to make a variable the subject

  1. For each of the following formulae, change the subject of the equation to the quantity indicated in the bracket on the right:
    1. \(t=\frac{v-u}{a}\)
    2. \(t=\sqrt{\frac{2s}{a}}\)
    3. \(u=\frac{s-\frac{1}{2}at^2}{t}\) or \(s=\frac{s}{t}-\frac{1}{2}at\)
    4. \(a=\frac{2(s-ut)}{t^2}\)
    5. \(s=\frac{v^2-u^2}{2a}\)
    6. \(u=\sqrt{v^2-2as}\)
    7. \(x=\left(\frac{y-a}{b}\right)^{\frac{1}{3}}\)
    8. \(\ln(\frac{i}{5})\) or \(\ln(i)-\ln(5)\)
    9. \(t=-\frac{1}{2}\ln(i/8)\) or \(-\frac{1}{2}\ln(i)+\frac{1}{2}\ln(8)\)
    10. \(x=\frac{1}{2}\left(\log_{10}-1\right)\)
    11. \(x=\frac{1}{3}\left(\log_{10}(y)+2\right)\)
    12. \(t=-\frac{1}{k}\ln\left(\frac{y-c_0}{a_0}\right)\)
  2. For each of the following, change the subject of the equation to the quantity indicated in the bracket on the right:
    1. \(y=\frac{2-x}{3+x}, \qquad \Longrightarrow x=\frac{2-3y}{y+1}\)
    2. \(y=\frac{4+2x}{5-x}, \qquad \Longrightarrow x=\frac{5y-4}{y+2}\)
    3. \(z=\frac{xy}{x+y}, \qquad \Longrightarrow y=\frac{xz}{x-z}\)
    4. \(C=\frac{C_1C_2}{C_1-C_2}, \qquad \Longrightarrow C_2=\frac{CC_1}{C+C_1}\)

9.5 Factorizing quadratics

  1. Factorize each of the following quadratic expressions:
  1. \(x^2+3x+2 =(x+1)(x+2)\)
  2. \(x^2+5x+4=(x+1)(x+4)\)
  3. \(x^2+4x+4=(x+2)^2\)
  4. \(x^2+x-2=(x+2)(x-1)\)
  5. \(x^2-x-2=(x-2)(x+1)\)
  6. \(x^2+5x-6=(x+6)(x-1)\)
  7. \(x^2-5x-6=(x-6)(x+1)\)
  8. \(x^2+x-6=(x+3)(x-2)\)
  9. \(x^2-x-6=(x-3)(x+2)\)
  10. \(2x^2+11x+12=(2x+3)(x+4)\)

9.6 Solving equations – variety of learned methods

  1. Solve the following equations: where answers are decimals, give accurate to five decimal places

    1. \(4x+5=8 \Longrightarrow x=\frac{3}{4}\)
    2. \(10x-8=-12\Longrightarrow x=-\frac{2}{5}\)
    3. \(6x+3=2x-5\Longrightarrow x=-2\)
    4. \(3x-9=5x+2\Longrightarrow x=-\frac{11}{2}\)
    5. \(5+3x^2=32\Longrightarrow x=3 \text{ and }x=-3\)
    6. \(5x^3+320=0\Longrightarrow x=4\)
    7. \(e^x=0.75\Longrightarrow x=\ln(0.75)=-0.28768\)
    8. \(e^{4x}=0.2\Longrightarrow x=\frac{1}{4}\ln(0.2)=-0.40236\)
    9. \(5e^{-x}-8=7\Longrightarrow -\ln(3)=-1.09861\)
    10. \(4+12e^{-3x}=13\Longrightarrow x=-\frac{1}{3}\ln(0.75)=0.09589\)
    11. \(\ln(3x)=-2\Longrightarrow x=\frac{1}{3}e^{-2}=0.04511\)
    12. \(5\ln(2x)+3=0\Longrightarrow x=\frac{1}{2}e^{-0.6}=0.27441\)
    13. \(5-3\ln(4x)=-10\Longrightarrow x=\frac{1}{4}e^5=37.10329\)
    14. \(\ln(6x+4)=0.25\Longrightarrow x=\frac{1}{6}\left(e^{0.25}-4\right)=-0.45266\)
    15. \(10^{x}=2.5\Longrightarrow x=\log_{10}(2.5)=0.39794\)
    16. \(10^{3x-2}=1.75\Longrightarrow x=\frac{1}{3}\left(\log_{10}(1.75)+2\right)=0.74768\)
    17. \(\log_{10}(4x)=0.32\Longrightarrow x=\frac{1}{4}10^{0.32}=0.52232\)
    18. \(\log_{10}(5x+4)=0.65\Longrightarrow x=\frac{1}{5}\left(10^{0.65}-4\right)=0.09337\)
    19. \(2^{4x-1}=5\Longrightarrow x=\frac{1}{4}\left(\frac{\log(5)}{\log(2)}+1 \right)=0.83048\)
    20. \(3^{2x+4}=6\Longrightarrow x=\frac{1}{2}\left(\frac{\log(6)}{\log(3)}-4 \right)=-1.18454\) any choice of log base is fine.

  2. Solve the following quadratic equations by factorization, then repeat using the quadratic formula and check your answers match.

    1. \(x^2+2x-15=(x+5)(x-3) \Longrightarrow x=-5 \text{ and } x=3\)
    2. \(x^2-9x+20=(x-5)(x-4) \Longrightarrow x=4 \text{ and } x=5\)
    3. \(x^2+10x+25=(x+5)^2 \Longrightarrow x=-5 \text{ (repeated)}\)
    4. \(2x^2-7x-4=(2x+1)(x-4) \Longrightarrow x=-\frac{1}{2} \text{ and } x=4\)

  3. Complete the square on each of these quadratic equations. Then try solving them using the quadratic formula, what happens?

    1. \(x^2 + 2x + 2 = (x+1)^2+1\). No solution possible.
    2. \(x^2 + 4x- 9=(x+2)^2-13\). Solutions are \(x=-2+\sqrt{13} \text{ and } x=-2-\sqrt{13}\).
    3. \(x^2+4x+9=(x+2)^2+5\). No solution possible.
    4. \(2x^2 -3x + 4 = 2((x-\frac{3}{4})^2+\frac{1}{4})=2(x-\frac{3}{4})^2+\frac{1}{4}\). No solution possible.

    In all cases where there was no (real) solution, the number outside the completed square bracket was positive. This means that if we tried to re-arrange to make \(x\) the subject we would need to square root a negative number, which we cannot do. This is because when you square a number the answer is never negative.

  4. Solve the following pairs of simultaneous equations:

    1. \[\begin{align*} 4x+3y&=2\\ 2x-y&=16 \end{align*}\] \(\Longrightarrow x=5, y=-6\)
    2. \[\begin{align*} 5x+2y&=1\\ 4x+3y&=-2 \end{align*}\] \(\Longrightarrow x=1, y=-2\)
    3. \[\begin{align*} 6x-2y&=-20\\ 4x+5y&=-7 \end{align*}\] \(\Longrightarrow x=-3, y=1\)
    4. \[\begin{align*} 8x-5y&=24.5\\ 2x-3y&=10.5 \end{align*}\] \(\Longrightarrow x=1.5, y=-2.5\)