DC CAPE Algebra II Creating Equations and Inequalities. Practice it free.

Subtract the third-year base from the first: $\text{\char36}35{,}200 - \text{\char36}31{,}000 = \text{\char36}4{,}200$. Over three years there are two equal increases, so divide by $3 - 1 = 2$ to get $x = \text{\char36}2{,}100$. Add $\text{\char36}2{,}100$ each year after the first: $\text{\char36}31{,}000$, $\text{\char36}33{,}100$, $\text{\char36}35{,}200$. Sum all three years: $\text{\char36}99{,}300$.

Set $y = 0$: $2x = 12$, so $x = 6$.

The slope is $m = \dfrac{5 - (-1)}{2 - (-4)} = 1$. Substitute $(-4, -1)$: $-1 = 1(-4) + b$, so $b = 3$. The equation is $y = x + 3$.

If the zeros are $x=2$ and $x=4$, that means the factors of the equation are $(x-2)$ and $(x-4)$. The product of these factors is $x^{2}-6x+8$. Comparing with $x^{2}+mx+8$, the missing linear coefficient is $m=-6$.

Cross-multiply to clear the fraction: $5(4x - 3y) = 2(x + y)$, so $20x - 15y = 2x + 2y$. Collect like terms to get $18x = 17y$. Divide both sides by $y$, then by $18$, to obtain $\dfrac{x}{y} = \dfrac{17}{18}\textsf{.}$

Factor the numerator as a perfect square: $x^{2} + 6x + 9 = (x+3)^{2}$. For $x \neq -3$, one factor of $x+3$ cancels, giving $x+3 = 4$, so $x = 1\textsf{.}$

The small jugs hold $2.5x$ gallons in all and the large jugs hold $5y$ gallons in all. The total volume is $47.5$ gallons, so $2.5x + 5y = 47.5$.

Above $32$ and below $40$ are both strict, so $32 < t < 40$.