Question

Solve the set of equations: $3(2u+v)=7uv$ and $3(u+3v)=11uv$

• A. $u=1;v=\frac{3}{2}$
• B. $u=2;v=\frac{1}{2}$
• C. $u=2;v=\frac{5}{4}$
• D. $u=5;v=\frac{7}{4}$

Answer 1

The correct option is A $u=1;v=\frac{3}{2}$

The equation $3(2u+v)=7uv$ can be rewritten as $6u+3v=7uv$ and the equation $3(u+3v)=11uv$ can be rewritten as $3u+9v=11uv$. Then we get the equations:

$6u+3v=7uv.........\left(1\right)$

$3u+9v=11uv.........\left(2\right)$

Multiply equation 2 by $2$:

$6u+18v=22uv.........\left(3\right)$

Subtract Equation 1 from equation 3 to eliminate $u$, because the coefficients of $u$ are the same. So, we get

$(6u-6u)+(18v-3v)=22uv-7uv$

i.e. $-15v=-15uv$

i.e. $u=1$

Substituting this value of $u$ in equation 1, we get

$6+3v=7v$

i.e. $4v=6$

i.e. $v=\frac{3}{2}$

Hence, the solution of the equations is $u=1,v=\frac{3}{2}$.