Substitution

Solve: \(3x + y = 15 and y = 2x\)

Solve: \(y = x + 5 and 2x - y = 0\)

Solve using substitution: \(y = 3x and x + y = 8\)

Solve: \(y = x - 4 and 3x + y = 8\)

Solve: \(x = 2y and 3x + y = 14\)

Solve: \(x = y + 5 and x = 2y\)

Solve: \(y = 2x - 7 and y = -x + 2\)

Solve: \(x = 2y + 4 and x + y = 1\)

Solve: \(x = y and 5x - y = 12\)

Solve: \(x = 5 and 2x + y = 13\)

Solve: \(y = 3x and x + y = 0\)

Solve: \(x = y - 2 and x + y = 10\)

Solve: \(y = -x + 4 and y = 2x - 2\)

Solve: \(x = y and 2x + y = 0\)

Solve: \(x = 3y and 2x + y = 14\)

Solve: \(x = y - 5 and 3x + y = 1\)

Solve: \(x = y + 7 and 2x + y = 8\)

Solve: \(2x + y = 7 and y = 3\)

Solve: \(y = 2x + 3 and y = 4x - 1\)

Solve: \(y = 4x and x + y = 5\)

Solve: \(y = x and x + 2y = 6\)

Solve: \(x = 2 and 4x + 3y = 11\)

Solve: \(x = y + 1 and x + y = 9\)

Solve: \(4x - 3y = 1 and x = y\)

Solve: \(y = x - 1 and y = -2x + 5\)

Solve: \(y = 4 and 3x - y = 5\)

Solve: \(y = 3x - 5 and y = 2x + 1\)

Solve: \(y = x + 2 and 2x + 2y = 12\)

Solve: \(y = 10 - x and 2x + y = 15\)

Solve: \(y = 2x and 4x - y = 10\)

Solve: \(y = 0.5x and x + y = 3\)

Solve: \(y = 2x - 3 and 5x - 2y = 10\)

If you are solving { x = y + 2 ; 3x + y = 10 } by substitution, what should the second equation look like after the first step?

Solve: \(y = x + 10 and y = 3x\)

Solve: \(3x - 2y = 12 and y = 0\)

Solve: \(y = x and 2x + y = 9\)

Solve: \(y = x + 1 and y = -x + 5\)

Solve: \(x = 3y - 1 and x + y = 11\)

For the system { y = 4x - 1 ; 2x + 3y = 10 }, which method is most efficient to use first?

Solve using substitution: \(y = 2x and x + y = 9.\)

Solve: \(x + 2y = 8 and x = 2y\)

Solve: \(x = 4 and x + y = 10\)

Solve: \(2x + 3y = 6 and x = 0\)

Solve: \(10x + y = 25 and y = 5\)

Solve: \(x = 12 - 3y and x + y = 6\)

Solve: \(3x + 2y = 12 and y = x - 4\)

Solve: \(y = 3x and 2x + y = 15\)

Solve: \(y = 2x + 1 and x + y = 7\)

Solve: \(3x + 2y = 13 and y = 2\)

Solve: \(x = -2 and x - 2y = 4\)