Exercise (5.5) : Solving Linear Quadratic Systems Algebraically



1.           Find the solution set of each of the systems of equations:

              (a)   $x^2-y^2=9$

                     $x+y=1$

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$\begin{array}{l}{{x}^{2}}-{{y}^{2}}=9\cdot \cdot \cdot \cdot \cdot \cdot \cdot (1)\\\\x+y=1\ \ \ \cdot \cdot \cdot \cdot \cdot \cdot \cdot (2)\\\\\text{From equation }(2)\text{,we have}\\\\y=1-x\ \ \ \cdot \cdot \cdot \cdot \cdot \cdot \cdot (3)\\\\\text{Substituting }y=1-x\text{ in equation }(1)\text{,}\\\\{{x}^{2}}-{{\left( {1-x} \right)}^{2}}=9\\\\{{x}^{2}}-\left( {1-2x+{{x}^{2}}} \right)=9\\\\{{x}^{2}}-1+2x-{{x}^{2}}=9\\\\2x=10\\\\x=5\\\\\text{Substituting }x=5\text{ in equation }(3),\\\\y=1-5=-4\\\\\therefore \ \ \text{Solution set}=\left\{ {\left( {5,-4} \right)} \right\}\text{ }\end{array}$

              (b)   $y=\displaystyle \frac{8}{x}$

                     $y=7+x$

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$\begin{array}{l}y=\displaystyle \frac{8}{x}\cdot \cdot \cdot \cdot \cdot \cdot \cdot (1)\\\\y=7+x\ \ \ \cdot \cdot \cdot \cdot \cdot \cdot \cdot (2)\\\\\text{From equation }(2)\text{,we have}\\\\\displaystyle \frac{8}{x}=7+x\\\\\therefore \ {{x}^{2}}+7x=8\\\\{{x}^{2}}+7x-8=0\\\\(x+8)(x-1)=0\\\\x=-8\ \text{or}\ x=1\\\\\text{When}\ x=-8,\ y=7-8=-1\\\\\text{When}\ x=1,\ y=7+1=8\\\\\therefore \ \ \text{Solution set}=\left\{ {\left( {-8,-1} \right),\left( {1,8} \right)} \right\}\text{ }\end{array}$

              (c)   $x^2+5x+y=4$

                     $x+y=8$

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$\begin{array}{l}{{x}^{2}}+5x+y=4\cdot \cdot \cdot \cdot \cdot \cdot \cdot (1)\\\\x+y=8\ \ \ \ \ \ \ \cdot \cdot \cdot \cdot \cdot \cdot \cdot (2)\\\\\text{From equation }(2)\text{,we have}\\\\y=8-x\ \ \ \ \ \ \cdot \cdot \cdot \cdot \cdot \cdot \cdot (3)\\\\\text{Substituting}\ y=8-x\text{ in equation }(1),\\\\{{x}^{2}}+5x+8-x=4\\\\{{x}^{2}}+4x+4=0\\\\{{(x+2)}^{2}}=0\\\\x=-2\\\\\text{Substituting}\ x=-2\text{ in equation }(3),\\\\y=8-(-2)=10\\\\\therefore \ \ \text{Solution set}=\left\{ {\left( {-2,10} \right)} \right\}\text{ }\end{array}$

2.           The sum of squares of two numbers is 58. If the first number and twice the second add up to 13, find the numbers.

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$\begin{array}{l}\text{Let}\ \text{the two numbers be }x\ \text{and }y.\\\\\text{By the problem,}\\\\{{x}^{2}}+{{y}^{2}}=58\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \ (1)\\\\x+2y=13\ \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (2)\\\\\text{From equation }(2)\text{,we have}\\\\x=13-2y\ \ \ \ \ \ \cdot \cdot \cdot \cdot \cdot \cdot \cdot (3)\\\\\text{Substituting}\ x=13-2y\text{ in equation }(1),\\\\{{\left( {13-2y} \right)}^{2}}+{{y}^{2}}=58\\\\169-52y+4{{y}^{2}}+{{y}^{2}}=58\\\\169-52y+5{{y}^{2}}=58\\\\\therefore \ \ 5{{y}^{2}}-52y+111=0\\\\(5y-37)(y-3)=0\\\\y=\displaystyle \frac{{37}}{5}\ \text{or}\ y=3\\\\\text{When}\ y=\displaystyle \frac{{37}}{5},x=13-2\left( {\displaystyle \frac{{37}}{5}} \right)\text{=}-\displaystyle \frac{9}{5}\text{ }\\\\\text{When}\ y=3,x=13-2\left( 3 \right)\text{=}7\text{ }\\\\\therefore \ \ \text{The two numbers are }-\displaystyle \frac{9}{5}\ \text{and }\displaystyle \frac{{37}}{5}\ \text{or 7 and 3}\text{.}\end{array}$

3.           The sum of the reciprocals of two positive numbers is $\displaystyle \frac{7}{36}$ and the product of the numbers is 108. Find the numbers.

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$\begin{array}{l}\text{Let}\ \text{the two positive numbers be }x\ \text{and }y.\\\\\text{By the problem,}\\\\\displaystyle \frac{1}{x}+\displaystyle \frac{1}{y}=\displaystyle \frac{7}{{36}}\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \ (1)\\\\xy=108\ \ \ \ \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (2)\\\\\text{From equation }(2)\text{,we have}\\\\y=\displaystyle \frac{{108}}{x}\ \ \ \ \ \ \cdot \cdot \cdot \cdot \cdot \cdot \cdot (3)\\\\\text{Substituting}\ y=\displaystyle \frac{{108}}{x}\text{ in equation }(1),\\\\\displaystyle \frac{1}{x}+\displaystyle \frac{1}{{\displaystyle \frac{{108}}{x}}}=\displaystyle \frac{7}{{36}}\\\\\displaystyle \frac{1}{x}+\displaystyle \frac{x}{{108}}=\displaystyle \frac{7}{{36}}\\\\\text{Multiplying both sides of equation by}\ 108x,\\\\108+{{x}^{2}}=21x\\\\\therefore \ \ {{x}^{2}}+21x-108=0\\\\(x-9)(x-12)=0\\\\x=9\ \text{or}\ x=12\\\\\text{When}\ x=9,y=\displaystyle \frac{{108}}{9}=12\\\\\text{When}\ x=12,y=\displaystyle \frac{{108}}{{12}}=9\\\\\therefore \ \ \text{The two positive numbers are }9\text{ and 12}\text{.}\end{array}$