Answer:
[tex]p=3[/tex] and [tex]q=4[/tex] the mistake was in calculating [tex]3q-4q-10=6[/tex] let's solve first.
Step-by-step explanation:
if [tex]p= 2q-5[/tex], substitute the [tex]p[/tex] in the equation[tex]3q-2p=6[/tex] as
[tex]3q-2(2q-5)=6[/tex]
solve the equation in parentheses :
[tex]3q-2(2q-5)\\[/tex]
multiply [tex]-2 \\[/tex] by [tex](2q-5)\\[/tex] it will equal to: [tex]3q-4q+10[/tex]
now we want to let all unknown digits alone, so we will take [tex]10[/tex] to the other side with a different sign it will be as:
[tex]3q-4q=6-10[/tex]
now we will solve it:
[tex]3q-4q=6-10\\q=-4[/tex] at this part, the given equation was wrong in the paper.
we don't want the digit to be negative so we will do it like this:
[tex]\frac{-q}{-1} =\frac{-4}{-1 } = q=4[/tex]
now that we found [tex]q[/tex] we will substitute it to find [tex]p\\[/tex]:
[tex]p=2q-5\\q=4\\p=2(4)-5\\p=8-5\\p=3[/tex]
overall the mistake was in calculating [tex]q[/tex] instead of putting [tex]3q-4q=6-10[/tex]
he/she put a different sign in front of the digit like :[tex]3q-4q-10=6[/tex]
so when he/she flipped [tex]-10[/tex] it became positive and instead of subtracting he/she added it.
