Question

please solve this.....​

Answer 1

GIVEN :–

$\sf \implies\left|\begin{array}{cc}\sf \: 2x& \sf5\\ \\ \sf \: 8 & \sf\:x\end{array}\right| = \left|\begin{array}{cc}\sf \: 6& \sf - 2\\ \\ \sf \:7 & \sf\:3\end{array}\right|$

TO FIND :–

• Value of 'x' = ?

SOLUTION :–

$\sf \implies\left|\begin{array}{cc}\sf \: 2x& \sf5\\ \\ \sf \: 8 & \sf\:x\end{array}\right| = \left|\begin{array}{cc}\sf \: 6& \sf - 2\\ \\ \sf \:7 & \sf\:3\end{array}\right|$

• We know that If –

$\sf \implies\left|\begin{array}{cc}\sf \: a& \sf \:b\\ \\ \sf \:c& \sf\:d\end{array}\right| = ad - bc$

• So that –

$\sf \implies(2x)(x) - (8)(5) = (6)(3) - ( - 7)(2)$

$\sf \implies2 {x}^{2} - 40 = 18 - ( - 14)$

$\sf \implies2 {x}^{2} - 40 = 18 + 14$

$\sf \implies2 {x}^{2} - 40 = 32$

$\sf \implies2 {x}^{2}=40 + 32$

$\sf \implies2 {x}^{2}=72$

$\sf \implies {x}^{2}= \cancel \dfrac{72}{2}$

$\sf \implies {x}^{2}= 36$

$\sf \implies x = \sqrt{36}$

$\sf \implies \large{ \boxed{ \sf x = \pm6}}$

Answer 2

Answer:

GIVEN :–

$\begin{gathered}\sf \implies|\begin{array}{cc}\sf \: 2x& \sf5\\ \\ \sf \: 8 & \sf\:x\end{array}| = |\begin{array}{cc}\sf \: 6& \sf - 2\\ \\ \sf \:7 & \sf\:3\end{array}|\end{gathered}$

TO FIND :–

• Value of 'x' = ?

SOLUTION :–

$\begin{gathered}\sf \implies|\begin{array}{cc}\sf \: 2x& \sf5\\ \\ \sf \: 8 & \sf\:x\end{array}| = |\begin{array}{cc}\sf \: 6& \sf - 2\\ \\ \sf \:7 & \sf\:3\end{array}|\end{gathered}$

• We know that If –

$\begin{gathered}\sf \implies|\begin{array}{cc}\sf \: a& \sf \:b\\ \\ \sf \:c& \sf\:d\end{array}| = ad - bc\end{gathered}$

• So that –

$\sf \implies\green{(2x)(x) - (8)(5) = (6)(3) - ( - 7)(2)}$

$\sf \implies\green{2 {x}^{2} - 40 = 18 - ( - 14)}$

$\sf \implies\green{2 {x}^{2} - 40 = 18 + 14}$

$\sf \implies\green{2 {x}^{2} - 40 = 32}$

$\sf \implies\green{2 {x}^{2}=40 + 32}$

$\sf \implies\green{2 {x}^{2}=72}$

$\sf \implies\green{ {x}^{2}= \cancel \dfrac{72}{2}}$

$\sf \implies\green{ {x}^{2}= 36}$

$\sf \implies\green{ x = \sqrt{36}}$

$\sf \implies \large{ \boxed{ \sf x = \pm6}}$