[tex]\sf \red{——————————————————————————}[/tex]
⸙Solve.
1.)
Total distance = 90 km
Increased speed = x+15
[tex] \tt \large \frac{ 90}{ \times } - \frac{90}{ \times } + 15 = \frac{1}{2} \\ \tt \large 90 \frac{1}{ \times } - \frac{1}{ \times } + 15 = \frac{1}{2} \\ \tt \large 90( \times + 15 - \times 2 + 15 \times ) = \frac{1}{2} [/tex]
[tex] \tt \large 90( \frac{ \times + 15 - 5}{ { \times }^{2} + \times } ) = \frac{1}{2} [/tex]
[tex]\large\boxed {\begin{array}{}{{\tt{In \: Equation:}}} \end{array}}[/tex]
[tex]\tt \large x²+15x-2700=0[/tex]
[tex]\tt \large x²+60x-45x-2700=0[/tex]
[tex]\tt \large x(x+60)-45(x+60)=0[/tex]
[tex]\tt \large (x+60) \: (x-45)=0[/tex]
[tex]\tt \large x=-60, x=45[/tex]
- The negative value must be cancelled = 60.
Therefore the usual speed is
[tex]{\underline{\boxed{\purple{\tt{45km/hr}}}}}[/tex]
2.)
[tex]\large\boxed {\begin{array}{}{{\tt{Solutions:}}} \end{array}}[/tex]
- Let x-be the number then,
[tex]\tt\large x- \frac{1}{ \times}= \frac{24}{5}[/tex]
- Using the least common denominator
[tex]\tt \large 5x[x- \frac{1}{ \times}= \frac{24}{5]}[/tex]
[tex]\tt \large (5x)(x)-(5)(1)=(24)(x)[/tex]
[tex]\tt \large 5x²-5=24x[/tex]
- Write the equation in standard form
[tex]\tt \large 5x²-5=24x[/tex]
[tex]\tt \large 5x²-24x-5=0[/tex]
[tex]\tt \large 5x²-24x-5=0[/tex]
[tex]\tt \large 5x²-24-5=0[/tex]
[tex]\tt \large (5x+1)(x-5)=0[/tex]
[tex]\tt \large 5x+=1x-5[/tex]
[tex]\tt \large 5x=1x=5[/tex]
[tex]\tt \large x= \frac{-1}{5}[/tex]
Therefore the answer is
[tex]{\underline{\boxed{\purple{\tt{5}}}}}[/tex]
[tex]\sf\red{——————————————————————————}[/tex]
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