Solve for y.

Question

1<y+ 10/3 ≤5
Write your answer as a compound inequality with integers.

Answer 1

To solve the compound inequality $1<y+\frac{10}{3}\le 5$, we'll break it down into two parts and solve them separately.

1. Solve the lower part of the inequality: $$1<y+\frac{10}{3}$$ Subtract $\frac{10}{3}$ from both sides: $$1-\frac{10}{3}<y$$ To combine the terms on the left, convert $1$ to a fraction with a denominator of $3$: $$1=\frac{3}{3}$$ Thus, we have: $$\frac{3}{3}-\frac{10}{3}<y$$ Simplifying this gives: $$\frac{3-10}{3}<y⟹\frac{-7}{3}<y$$

2. Solve the upper part of the inequality: $$y+\frac{10}{3}\le 5$$ Subtract $\frac{10}{3}$ from both sides: $$y\le 5-\frac{10}{3}$$ Convert $5$ to a fraction with a denominator of $3$: $$5=\frac{15}{3}$$ Therefore, we have: $$y\le \frac{15}{3}-\frac{10}{3}$$ Simplifying this gives: $$y\le \frac{15-10}{3}⟹y\le \frac{5}{3}$$

Now, we combine both parts of the inequality: $$\frac{-7}{3}<y\le \frac{5}{3}$$

To express the solution with integers, we can convert the fractional bounds to decimal equivalents: $$-2.33<y\le 1.67$$

The integers that fall within this range are $-2,-1,0,1$.

Thus, the solution in integer form is: $$y\in -2,-1,0,1$$

Or as a compound inequality with integers: $$-2\le y\le 1$$