Which of these equations is linear in \(x\) and \(y\): \(3x-2y=7\), \(xy=4\), or \(x^2+y=1\)? Explain your choice.

For the system \(2x+y=5\), \(x-y=1\), name the coefficient matrix \(A\), unknown vector \(\mathbf{x}\), and right-hand-side vector \(\mathbf{b}\) in \(A\mathbf{x}=\mathbf{b}\).

Write the equations represented by \(\begin{pmatrix}4&-1\\0&3\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}6\\9\end{pmatrix}\).

A system is written as \(y+2x=8\) and \(3x-y=1\). Rewrite it in the consistent variable order \((x,y)\).

Solve the system \(x+y=9\), \(x-y=3\) by elimination.

Solve the system \(2x+y=11\), \(x+y=7\).

Solve \(3x+2y=16\), \(x+2y=8\), choosing an elimination step that removes one variable quickly.

Solve \(2x-3y=1\), \(4x+y=17\).

Solve the three-variable system \(x+y+z=6\), \(x-y+z=2\), \(2x+z=7\).

Solve \(x+2y-z=4\), \(2x-y+z=1\), \(x+y+z=6\).

The system \(x+y=4\), \(2x+2y=8\) is solved by letting \(y=t\). Write the full solution set and explain why there are infinitely many solutions.

Check whether \((x,y)=(2,5)\) is a solution of the system \(3x-y=1\), \(x+y=7\), and then solve the system.

For what value of \(k\) does the system \(x+y=3\), \(2x+2y=k\) have at least one solution? Describe the solutions for that value.

A particle's constant velocity components \((u,v)\) satisfy \(2u+v=13\) and \(u-v=2\). Solve for \((u,v)\) and check both equations.

For which values of \(a\) does \(x+y=2\), \(2x+2y=a\) have no solution, and for which value does it have infinitely many solutions?

Find all values of \(a\) for which \(ax+y=1\), \(2x+2y=2\) has infinitely many solutions, one solution, or no solution.

The system \(x+y+z=5\), \(2x+2y+2z=10\), \(x-y=1\) has a free variable. Write its solution set.

A learner claims that multiplying one equation in a system by \(0\) is a valid elimination step because \(0=0\) is always true. Explain the error using \(x+y=3\), \(x-y=1\).

Prove briefly that swapping the order of equations in a linear system cannot change the solution set.

Diagnose the mistake: from \(x+y=6\) and \(x-y=2\), a learner subtracts and writes \(2y=4\), then concludes \(y=2\) and \(x=4\). Is the method sound?