[tex]\bf \begin{cases} 2p+q=11\\ p+2q=13 \end{cases}\implies (\stackrel{11}{2p+q})+(\stackrel{13}{p+2q})=\stackrel{11+13}{24} \\\\\\ 3p+3q=24\implies 3(p+q)=24\implies p+q=\cfrac{24}{3}\implies p+q=8[/tex]
Answer:
the desired result is (p + q) = (3 + 5) = 8
Step-by-step explanation:
Let's solve this system of linear equations in the usual way: find the values of p and q. Then find (p + q) as a numerical result.
Solve:
2p+q=11
p+2q=13
Multiply the second equation by -2:
2p + q = 11
-2p - 4q = -26
Combining these two equations results in -3q = -15, and so q must be 5.
Subbing 5 for q in the first equation, we get:
2p + 5 = 11, or 2p = 6. Then p = 3.
Then the desired result is (p + q) = (3 + 5) = 8
