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.A wooden block of mass M lies on a rough floor. Another wooden block of the same mass is hanging from the point O through strings as shown in the figure. To achieve equilibrium, the coefficient of static friction between the block on the floor and the floor itself is
In an experiment to determine the figure of merit of a galvanometer by half deflection method, a student constructed the following circuit. He applied a resistance of \( 520 \, \Omega \) in \( R \). When \( K_1 \) is closed and \( K_2 \) is open, the deflection observed in the galvanometer is 20 div. When \( K_1 \) is also closed and a resistance of \( 90 \, \Omega \) is removed in \( S \), the deflection becomes 13 div. The resistance of galvanometer is nearly: