Question

Solve for x and y
4$\left[\begin{array}{ccc}1&\\2x\end{array}\right] + \left[\begin{array}{ccc}4x\\3y\end{array}\right] = \left[\begin{array}{ccc}16\\18\end{array}\right]$

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{4}\left[\begin{array}{c}\tt{1}&\tt{2x}\end{array}\right] +\left[\begin{array}{c}\tt{4x}&\tt{3y}\end{array}\right] =\left[\begin{array}{c}\tt{16}&\tt{18}\end{array}\right]$

$\implies\left[\begin{array}{c}\sf{4}&\sf{8x}\end{array}\right] +\left[\begin{array}{c}\sf{4x}&\sf{3y}\end{array}\right] =\left[\begin{array}{c}\sf{16}&\sf{18}\end{array}\right]$

$\implies\left[\begin{array}{c}\sf{4+4x}&\sf{8x+3y}\end{array}\right] =\left[\begin{array}{c}\sf{16}&\sf{18}\end{array}\right]$

$\sf{4+4x=16\,\,\,\,\,....(i)}\\\sf{8x+3y=18\,\,\,\,\,....(ii)}$

Solving the equation (i),

$\sf{4(1+x)=16}$

$\sf{\implies1+x=4}$

$\sf{\implies\,x=3}$

Put this value in equation(ii),

$\sf{8(3)+3y=18}$

$\sf{\implies24+3y=18}$

$\sf{\implies3y=-6}$

$\sf{\implies\,y=-2}$

$\bf{\blue{Hence,\,\,required\,\,values\,\,of\,\, x\,\,and\,\,y\,\,are\,\,3\,\,and\,\,-2}}$