Question

Complete the following activity to solve the equations 3x – 5y = 11

and 2x + 7y = - 3

i) Find D, Dx and Dy from the equations

ii) Find the value of x and y using cramers rule.​

Answer 1

Answer:

i) $\displaystyle{{\boxed{\red{\sf\:D\:=\:31}}}}$

$\displaystyle{{\boxed{\green{\sf\:D_x\:=\:62}}}}$

$\displaystyle{{\boxed{\blue{\sf\:D_y\:=\:-\:31}}}}$

ii) $\displaystyle{\boxed{\red{\sf\:(\:x\:,\:y\:)\:=\:(\:2\:,\:-\:1\:)\:}}}$

Step-by-step-explanation:

i)

The given simultaneous equations are

3x - 5y = 11 - - - ( 1 ) &

2x + 7y = - 3 - - - ( 2 )

Now,

3x - 5y = 11 - - - ( 1 )

Comparing with ax + by = c, we get,

$\displaystyle{\sf\:\bullet\:a_1\:=\:3}$

$\displaystyle{\sf\:\bullet\:b_1\:=\:-\:5}$

$\displaystyle{\sf\:\bullet\:c_1\:=\:11}$

Now,

2x + 7y = - 3 - - - ( 2 )

Comparing with ax + by = c, we get,

$\displaystyle{\sf\:\bullet\:a_2\:=\:2}$

$\displaystyle{\sf\:\bullet\:b_2\:=\:7}$

$\displaystyle{\sf\:\bullet\:c_2\:=\:-\:3}$

Now, we know that,

$\displaystyle{\pink{\sf\:D\:=\:\left|\:\begin{array}{cc}\sf\:a_1 & \sf\:b_1\\\sf\:a_2 & \sf\:b_2\:\end{array}\right|}}$

$\displaystyle{\implies\sf\:D\:=\:\left|\begin{array}{cc}\sf\:3 & \sf\:-\:5\\\sf\:2 & \sf\:7\end{array}\right|}}$

$\displaystyle{\implies\sf\:D\:=\:(\:3\:\times\:7\:)\:-\:[\:(\:-\:5\:)\:\times\:2\:]}$

$\displaystyle{\implies\sf\:D\:=\:21\:-\:(\:-\:10\:)}$

$\displaystyle{\implies\sf\:D\:=\:21\:+\:10}$

$\displaystyle{\implies\underline{\boxed{\red{\sf\:D\:=\:31}}}}$

Now, we know that,

$\displaystyle{\pink{\sf\:D_x\:=\:\left|\:\begin{array}{cc}\sf\:c_1 & \sf\:b_1\\\sf\:c_2 & \sf\:b_2\:\end{array}\right|}}$

$\displaystyle{\implies\sf\:D_x\:=\:\left|\begin{array}{cc}\sf\:11 & \sf\:-\:5\\\sf\:-\:3 & \sf\:7\end{array}\right|}}$

$\displaystyle{\implies\sf\:D_x\:=\:(\:11\:\times\:7\:)\:-\:[\:(\:-\:5\:)\:\times\:(\:-\:3\:)\:]}$

$\displaystyle{\implies\sf\:D_x\:=\:77\:-\:(\:15\:)}$

$\displaystyle{\implies\sf\:D_x\:=\:77\:-\:15}$

$\displaystyle{\implies\underline{\boxed{\green{\sf\:D_x\:=\:62}}}}$

Now, we know that,

$\displaystyle{\pink{\sf\:D_y\:=\:\left|\:\begin{array}{cc}\sf\:a_1 & \sf\:c_1\\\sf\:a_2 & \sf\:c_2\:\end{array}\right|}}$

$\displaystyle{\implies\sf\:D_y\:=\:\left|\begin{array}{cc}\sf\:3 & \sf\:11\\\sf\:2 & \sf\:-\:3\end{array}\right|}}$

$\displaystyle{\implies\sf\:D_y\:=\:3\:\times\:(\:-\:3\:)\:-\:(\:11\:\times\:2\:)}$

$\displaystyle{\implies\sf\:D_y\:=\:-\:9\:-\:22}$

$\displaystyle{\implies\underline{\boxed{\blue{\sf\:D_y\:=\:-\:31}}}}$

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ii)

Now, we know that,

$\displaystyle{\pink{\sf\:x\:=\:\dfrac{D_x}{D}}\sf\:\quad\:-\:-\:-\:[\:Cramer's\:rule\:]}$

$\displaystyle{\implies\sf\:x\:=\:\cancel{\dfrac{62}{31}}}$

$\displaystyle{\implies\underline{\boxed{\red{\sf\:x\:=\:2}}}}$

Now,

$\displaystyle{\pink{\sf\:y\:=\:\dfrac{D_y}{D}}\sf\:\quad\:-\:-\:-\:[\:Cramer's\:rule\:]}$

$\displaystyle{\implies\sf\:y\:=\:\cancel{-\:\dfrac{31}{31}}}$

$\displaystyle{\implies\underline{\boxed{\blue{\sf\:y\:=\:-\:1}}}}$

∴ The values of x and y are 2 & - 1 respectively.