Question

after the addition of an acid, a solution has a volume of 90 milliliters. the volume of the solution is 3 milliliters greater than 3 times the volume of the solution before the acid was added. what was the original volume of the solution?

Explanation:

Step1: Let the original volume of the solution be $x$ milliliters.

Let the volume of the added acid be $y$ milliliters.

Step2: Set up equations based on given information.

We know that $x + y=90$ (total volume after addition is 90 milliliters) and $y=x + 3$ (volume of acid is 3 milliliters greater than original volume of solution).

Step3: Substitute the second - equation into the first one.

Substitute $y=x + 3$ into $x + y=90$, we get $x+(x + 3)=90$.

Step4: Simplify the equation.

Expand the left - hand side: $x+x + 3=90$, which simplifies to $2x+3 = 90$.

Step5: Solve for $x$.

Subtract 3 from both sides: $2x=90 - 3=87$. Then divide both sides by 2: $x=\frac{87}{2}=43.5$.

Answer:

43.5 milliliters