When solving systems of linear equations, express the system as separate equations for each variable. You can use substitution or elimination to solve for the unknowns. In this case, we used substitution to solve for \(x\) and \(y\).
The correct answer is: (B) x = 4, y = 3.
We are given the matrix equation:
\( x \begin{bmatrix} 3 \\ 2 \end{bmatrix} + y \begin{bmatrix} 1 \\ -1 \end{bmatrix} = \begin{bmatrix} 15 \\ 5 \end{bmatrix} \)
This represents a system of two linear equations in \(x\) and \(y\). We can express this equation as two separate equations:\( 3x + y = 15 \)
\( 2x - y = 5 \)
Now, solve the system of equations: - From the first equation, \( 3x + y = 15 \), we can solve for \(y\):\( y = 15 - 3x \)
- Substitute \(y = 15 - 3x\) into the second equation \(2x - y = 5\):\( 2x - (15 - 3x) = 5 \)
Simplifying this:\( 2x - 15 + 3x = 5 \)
\( 5x = 20 \)
Solving for \(x\):\( x = 4 \)
- Now substitute \(x = 4\) into \(y = 15 - 3x\):\( y = 15 - 3(4) = 15 - 12 = 3 \)
Therefore, the values of \(x\) and \(y\) are x = 4 and y = 3. Thus, the correct answer is (B) x = 4, y = 3.You are given a dipole of charge \( +q \) and \( -q \) separated by a distance \( 2l \). A sphere 'A' of radius \( R \) passes through the centre of the dipole as shown below and another sphere 'B' of radius \( 2R \) passes through the charge \( +q \). Then the electric flux through the sphere A is