Giải:
a) Có: \(a+b=18\Leftrightarrow a=18-b\)
Lại có: \(a-b=12\)
\(\Leftrightarrow18-b-b=12\)
\(\Leftrightarrow18-2b=12\)
\(\Leftrightarrow2b=18-12=6\)
\(\Leftrightarrow b=3\)
\(\Leftrightarrow a=18-b=18-3=15\)
Vậy ...
b) Có: \(a+b=50\Leftrightarrow a=50-b\)
Lại có: \(2a+5b=80\)
\(\Leftrightarrow2\left(50-b\right)+5b=80\)
\(\Leftrightarrow100-2b+5b=80\)
\(\Leftrightarrow100+3b=80\)
\(\Leftrightarrow3b=-20\)
\(\Leftrightarrow b=-\dfrac{20}{3}\)
\(\Leftrightarrow a=50-b=50-\left(-\dfrac{20}{3}\right)=\dfrac{170}{3}\)
Vậy ...
a) Ta có
\(a+b=18\)
\(\Rightarrow a=18-b\)
Và \(a-b=12\)
\(\Rightarrow18-b-b=12\)
\(\Rightarrow18-2b=12\)
\(\rightarrow2b=18-12=6\)
\(b=6:2=3\)
b)Ta có
\(a+b=50\)
\(\Rightarrow a=50-b\)
Và \(2a+5b=80\)
\(\Rightarrow2.\left(50-b\right)\)\(+5b=80\)
\(\Rightarrow100-2b+5b=80\)
\(\Rightarrow100\)\(+3b\)\(=80\)
\(\Rightarrow3b=-20\)
Vậy b=\(-20:3\)
\(\Rightarrow a=50-\dfrac{-20}{3}=\dfrac{170}{3}\)
