Answer Key 8.7
- [latex]\text{LCD}=2(x)[/latex]
[latex]\begin{array}[t]{rrcrrrl} 3x(2x)&-&x&-&2&=&0 \\ 6x^2&-&x&-&2&=&0 \\ (3x&-&2)(2x&+&1)&=&0 \\ \\ &&&&x&=&\dfrac{2}{3}, -\dfrac{1}{2} \end{array}[/latex] - [latex]\text{LCD}=x+1[/latex]
[latex]\begin{array}[t]{rrcrrrl} (x&+&1)(x&+&1)&=&\phantom{-}4 \\ x^2&+&2x&+&1&=&\phantom{-}4 \\ &&&-&4&&-4 \\ \hline x^2&+&2x&-&3&=&0 \\ (x&-&1)(x&+&3)&=&0 \\ \\ &&&&x&=&1, -3 \end{array}[/latex] - [latex]\text{LCD}=x-4[/latex]
[latex]\begin{array}[t]{rrcrrrllrrr} x(x&-&4)&+&20&=&5x&-&2(x&-&4) \\ x^2&-&4x&+&20&=&5x&-&2x&+&8 \\ &-&3x&-&8&&&-&3x&-&8 \\ \hline x^2&-&7x&+&12&=&0&&&& \\ (x&-&4)(x&-&3)&=&0&&&& \\ \\ &&&&x&=&3,&4&&& \end{array}[/latex] - [latex]\text{LCD}=x-1[/latex]
[latex]\begin{array}[t]{rrrrrrrrllr} x^2&+&6&+&x&-&2&=&\phantom{-}2x(x&-&1) \\ &&x^2&+&x&+&4&=&\phantom{-}2x^2&-&2x \\ &-&2x^2&+&2x&&&&-2x^2&+&2x \\ \hline &&-x^2&+&3x&+&4&=&0&& \\ &&x^2&-&3x&-&4&=&0&& \\ &&(x&-&4)(x&+&1)&=&0&& \\ \\ &&&&&&x&=&4, 1&& \\ \end{array}[/latex] - [latex]\text{LCD}=x-3[/latex]
[latex]\begin{array}[t]{rrcrrrr} x(x&-&3)&+&6&=&2x \\ x^2&-&3x&+&6&=&2x \\ &-&2x&&&&-2x \\ \hline x^2&-&5x&+&6&=&0 \\ (x&-&3)(x&-&2)&=&0 \\ \\ &&&&x&=&2, 3 \end{array}[/latex] - [latex]\text{LCD}=(x-1)(3-x)[/latex]
[latex]\begin{array}[t]{rrcrrrlrrrrrcrr} (x&-&4)(3&-&x)&=&\phantom{-}12(x&-&1)&+&(x&-&1)(3&-&x) \\ -x^2&+&7x&-&12&=&\phantom{-}12x&-&12&-&x^2&+&4x&-&3 \\ +x^2&-&16x&+&15&&-12x&+&12&+&x^2&-&4x&+&3 \\ \hline &&-9x&+&3&=&0&&&&&&&& \\ &&&&3&=&9x&&&&&&&& \\ \\ &&&&x&=&\dfrac{3}{9}\hspace{0.1in}\text{ or}&\dfrac{1}{3}&&&&&&& \end{array}[/latex] - [latex]\text{LCD}=(2m-5)(3m+1)(2)[/latex]
[latex]\begin{array}[t]{rrcrcrrrrrcrr} 3m(3m&+&1)(2)&-&7(2m&-&5)(2)&=&3(2m&-&5)(3m&+&1) \\ 18m^2&+&6m&-&28m&+&70&=&18m^2&-&39m&-&15 \\ -18m^2&&&+&39m&+&15&&-18m^2&+&39m&+&15 \\ \hline &&&&17m&+&85&=&0&&&& \\ &&&&&-&85&&-85&&&& \\ \hline &&&&&&\dfrac{17m}{17}&=&\dfrac{-85}{17}&&&& \\ \\ &&&&&&m&=&-5&&&& \end{array}[/latex] - [latex]\text{LCD}=(1-x)(3-x)[/latex]
[latex]\begin{array}[t]{rrcrrrrrr} (4&-&x)(3&-&x)&=&12(1&-&x) \\ 12&-&7x&+&x^2&=&12&-&12x \\ -12&+&12x&&&&-12&+&12x \\ \hline &&x^2&+&5x&=&0&& \\ &&x(x&+&5)&=&0&& \\ \\ &&&&x&=&0,&-5& \end{array}[/latex] - [latex]\text{LCD}=2(y-3)(y-4)[/latex]
[latex]\begin{array}[t]{crrrrrcrrrrrcrr} 7(2)(y&-&4)&-&1(y&-&3)(y&-&4)&=&(y&-&2)(2)(y&-&3) \\ 14y&-&56&-&y^2&+&7y&-&12&=&2y^2&-&10y&+&12 \\ \\ &&&&-\phantom{0}y^2&+&21y&-&68&=&2y^2&-&10y&+&12 \\ &&&&-2y^2&+&10y&-&12&&-2y^2&+&10y&-&12 \\ \hline &&&&-3y^2&+&31y&-&80&=&0&&&& \\ &&&&3y^2&-&31y&+&80&=&0&&&& \\ &&&&(y&-&5)(3y&-&16)&=&0&&&& \\ \\ &&&&&&&&y&=&5, &\dfrac{16}{3}&&& \end{array}[/latex] - [latex]\text{LCD}=(x+2)(x-2)[/latex]
[latex]\begin{array}[t]{rrrrrrrrrrr} 1(x&-&2)&+&1(x&+&2)&=&3x&+&8 \\ x&-&2&+&x&+&2&=&3x&+&8 \\ &&&&-2x&&&&-2x&& \\ \hline &&&&&&0&=&x&+&8 \\ &&&&&&-8&&&-&8 \\ \hline &&&&&&x&=&-8&& \end{array}[/latex] - [latex]\text{LCD}=(x+1)(x-1)(6)[/latex]
[latex]\begin{array}[t]{rrcrcrrrcrcrrrcrr} (x&+&1)(x&+&1)(6)&-&(x&-&1)(x&-&1)(6)&=&5(x&+&1)(x&-&1) \\ 6(x^2&+&2x&+&1)&-&6(x^2&-&2x&+&1)&=&5(x^2&&-&&1) \\ 6x^2&+&12x&+&6&-&6x^2&+&12x&-&6&=&5x^2&&&-&5 \\ &&&&&&&&&&24x&=&5x^2&&&-&5 \\ &&&&&&&&&&-24x&&&-&24x&& \\ \hline &&&&&&&&&&0&=&5x^2&-&24x&-&5 \\ &&&&&&&&&&0&=&(5x&+&1)(x&-&5) \\ \\ &&&&&&&&&&x&=&5, &-\dfrac{1}{5}&&& \end{array}[/latex] - [latex]\text{LCD}=(x+3)(x-2)[/latex]
[latex]\begin{array}[t]{rrcrcrrrrrr} (x&-&2)(x&-&2)&-&1(x&+&3)&=&1 \\ x^2&-&4x&+&4&-&x&-&3&=&1 \\ &&&&&&&-&1&&-1 \\ \hline &&&&&&x^2&-&5x&=&0 \\ &&&&&&x(x&-&5)&=&0 \\ \\ &&&&&&&&x&=&0, 5 \end{array}[/latex] - [latex]\text{LCD}=(x-1)(x+1)[/latex]
[latex]\begin{array}[t]{rrrrcrrrrrcrr} x(x&+&1)&-&2(x&-&1)&=&4x^2&&&& \\ x^2&+&x&-&2x&+&2&=&4x^2&&&& \\ -x^2&&&+&x&-&2&&-x^2&+&x&-&2 \\ \hline &&&&&&0&=&3x^2&+&x&-&2 \\ &&&&&&0&=&(3x&-&2)(x&+&1) \\ \\ &&&&&&0&=&\dfrac{2}{3},&-1&&& \end{array}[/latex] - [latex]\text{LCD}=(x+2)(x-4)[/latex]
[latex]\begin{array}[t]{rrrrcrrrr} 2x(x&-&4)&+&2(x&+&2)&=&3x \\ 2x^2&-&8x&+&2x&+&4&=&3x \\ &&&-&3x&&&&-3x \\ \hline &&2x^2&-&9x&+&4&=&0 \\ &&(2x&-&1)(x&-&4)&=&0 \\ \\ &&&&&&x&=&\dfrac{1}{2}, 4 \end{array}[/latex] - [latex]\text{LCD}=(x+1)(x+5)[/latex]
[latex]\begin{array}[t]{rrrrcrrrl} 2x(x&+&5)&-&3(x&+&1)&=&-8x^2 \\ 2x^2&+&10x&-&3x&-&3&=&-8x^2 \\ +8x^2&&&&&&&&+8x^2 \\ \hline &&10x^2&+&7x&-&3&=&0 \\ &&(10x&-&3)(x&+&1)&=&0 \\ \\ &&&&&&x&=&\dfrac{3}{10}, -1 \end{array}[/latex]
