In Exercises 91–100, find all values of x satisfying the given conditions.y1=6(2xx−3)2,y2=5(2xx−3),andy1 exceeds y2 by 6.$$y_1 = 6$$ \left( \frac{2x}{x - 3} \$$right)^2$$, \quad $$y_2 = 5$$ \left( \frac{2x}{x - 3} \right), \quad \text{and} \quad $$y_1$$ \text{ exceeds } $$y_2$$ \text{ by } 6. y1=6(x−32x)2,y2=5(x−32x),andy1 exceeds y2 by 6.

Ch. 1 - Equations and Inequalities

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Blitzer 8th Edition

Ch. 1 - Equations and Inequalities

Problema 100

Capitolo 2, Problema 100

In Exercises 91–100, find all values of x satisfying the given conditions. $y_1 = 6 $\left$( $\frac{2x}{x - 3}$ $\right$)^2, $\quad$ y_2 = 5 $\left$( $\frac{2x}{x - 3}$ $\right$), $\quad$ $\text{and}$ $\quad$ y_1 $\text{ exceeds }$ y_2 $\text{ by }$ 6.$

Guida verificata passo dopo passo

Start by translating the condition "y1 exceeds y2 by 6" into an equation. This means that y1 is equal to y2 plus 6, so write: $y_1 = y_2 + 6$.

Substitute the given expressions for $y_1$ and $y_2$ into the equation: $6\left(\frac{2x}{x - 3}\right)^2 = 5\left(\frac{2x}{x - 3}\right) + 6$.

To simplify the equation, let $t = \frac{2x}{x - 3}$. Rewrite the equation in terms of $t$: $6t^2 = 5t + 6$.

Rearrange the equation to standard quadratic form: $6t^2 - 5t - 6 = 0$.

Solve the quadratic equation for $t$ using the quadratic formula: $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a=6$, $b=-5$, and $c=-6$. After finding the values of $t$, substitute back $t = \frac{2x}{x - 3}$ and solve for $x$.

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