Question 531088
A KMST (chemist) is the right person to answer this question.
We have to assume that volumes are additive in this case, meaning that the volume after mixing will equal the sum of the volumes before mixing. (Not always true).
We also have to assume that the concentrations were given in percentage weight in volume (grams of acid per milliliter of solution). (Not usually the case, but done in some cases).
I want to assume that it is the same acid, because we do not like to add apples and oranges, and we know that millions of acids exist.

Let s be the volume (in milliliters) of 75% solution (strong solution) to be used.
Let w be the volume (in milliliters) of 50% solution (weak solution) to be used.
The amount of acid in each volume used will be 0.75s and 0.50w.
The amount of acid in the final mixture has to be {{{10*0.60=6}}}
That leads us to the equations
{{{s+w=10}}} and
{{{0.75s+0.5w=6}}}

If studying systems of linear equations, solve as you are being taught to do.
If not there yet, change {{{s+w=10}}} to {{{w=10-s}}}
substitute in the other equation to get
{{{0.75s+0.5(10-s)=6}}},
{{{0.75s+5-0.5s=6}}} and
{{{0.25s=1}}} and {{{s=4}}}
Then find {{{w}}}