Free Math Worksheets

- Divide the first equation by \(4\). - Compare the left sides of the two equations. - When are two lines identical, and when are they parallel? - Could lines with the same slope intersect exactly once?

1. Divide equation (I) by \(4\): \(3x - y = 2\). 2. When \(k = 2\), equation (II) is identical to equation (I), so the system has infinitely many solutions. 3. When \(k \ne 2\), the equations have identical left sides but different constants. They represent distinct parallel lines, so the system has no solution. 4. For every value of \(k\), the lines have the same slope. They are therefore either identical or parallel and can never intersect at exactly one point.

a) \(k = 2\) b) \(k \ne 2\) c) The lines always have the same slope, so they are either identical or distinct parallel lines.

- When do two equations represent the same line, and when do they represent parallel lines? - Compare corresponding coefficients and constants. - Equal or proportional left sides with nonproportional constants indicate no solution. - To create an equation through a point, substitute its coordinates into a simple linear form.

1. Divide equation (I) by \(2\): \(2x - 3y = 6\). Therefore, when \(k = 6\), equations (I) and (II) represent the same line and the system has infinitely many solutions. 2. When \(k = 10\), equation (II) is \(2x - 3y = 10\). Multiplying by \(2\) gives \(4x - 6y = 20\). This line has the same left side as equation (I) but a different constant, so the lines are parallel and the system has no solution. 3. A valid equation through \((3, 0)\) that is not a multiple of equation (I) is \(x + y = 3\). Together with equation (I), it has the unique solution \((3, 0)\).

a) \(k = 6\) b) The equations represent distinct parallel lines when \(k = 10\), so there is no solution. c) One possible equation is \(x + y = 3\).

- Solve the second equation for one variable and substitute. - Identify the expression that would be used as a divisor. - What happens when that expression equals zero?

1. Solve equation (II) for \(y\): \(y = 2x - 3\). 2. Substitute into equation (I): \(ax + 5(2x - 3) = 15\). 3. Simplify: \((a + 10)x = 30\). 4. If \(a \ne -10\), divide to get \(x = \frac{30}{a + 10}\). 5. Then \(y = 2x - 3 = \frac{30 - 3a}{a + 10}\). 6. If \(a = -10\), the equation becomes \(0 = 30\), so the system has no solution.

For \(a \ne -10\), \((x, y) = \left(\frac{30}{a + 10}, \frac{30 - 3a}{a + 10}\right)\). For \(a = -10\), the system has no solution.

- Scale the first equation so that the \(y\)-coefficients match. - What happens when the \(x\)-coefficients also match? - If they do not match, what must \(x\) equal?

1. Multiply equation (I) by \(3\): \(3x + 6y = 18\). 2. Subtract this equation from equation (II): \((m - 3)x = 0\). 3. If \(m = 3\), the original equations are equivalent, so the system has infinitely many solutions. 4. If \(m \ne 3\), then \(x = 0\). Substitute into equation (I): \(2y = 6\), so \(y = 3\). Thus, the unique solution is \((0, 3)\).

If \(m = 3\), the system has infinitely many solutions. If \(m \ne 3\), the system has the unique solution \((0, 3)\).

- Use substitution and note when a variable disappears. - When does the resulting equation become a contradiction? - What would be required for an identity?

1. Solve equation (II) for \(x\): \(x = 3y + 5\). 2. Substitute into equation (I): \(2(3y + 5) + ay = 8\), so \((a + 6)y = -2\). 3. If \(a = -6\), the equation becomes \(0 = -2\), a contradiction. Therefore, the system has no solution. 4. Infinitely many solutions would require an identity such as \(0 = 0\). Because the right side remains \(-2\), no value of \(a\) produces infinitely many solutions.

The system has no solution when \(a = -6\). There is no value of \(a\) for which it has infinitely many solutions.

- Scale one equation so the \(y\)-coefficients match. - When do matching variable coefficients create a contradiction? - When do they create equivalent equations? - What slope relationship guarantees one intersection?

1. Multiply equation (I) by \(-2\): \(-4x + 6y = -12\). 2. If \(a = -4\), the variable coefficients match. 3. For no solution, the constants must differ: \(a = -4\) and \(b \ne -12\). 4. For infinitely many solutions, the equations must be identical: \(a = -4\) and \(b = -12\). 5. For exactly one solution, the slopes must differ: \(a \ne -4\), with any value of \(b\).

1. \(a = -4\), \(b \ne -12\) 2. \(a = -4\), \(b = -12\) 3. \(a \ne -4\), with any \(b\)

- Eliminate one variable while keeping \(k\) in the coefficients. - For which values of \(k\) does the resulting variable coefficient become zero? - Substitute each resulting value of \(k\) into the elimination equation. - Does elimination produce an identity or a contradiction?

1. Multiply equation (I) by \(k + 1\): \(2(k + 1)x + (k^2 - 1)y = 6(k + 1)\). 2. Multiply equation (II) by \(2\): \(2(k + 1)x + 8y = 24\). 3. Subtract the second transformed equation from the first: \((k^2 - 9)y = 6k - 18\). 4. The system can fail to have exactly one solution only when the coefficient of \(y\) is zero: \(k^2 - 9 = 0\). Thus, \(k = 3\) or \(k = -3\). 5. If \(k = 3\), the elimination equation is \(0 = 0\). The original equations are equivalent, so there are infinitely many solutions. 6. If \(k = -3\), the elimination equation is \(0 = -36\), a contradiction, so there is no solution.

For \(k = 3\), the system has infinitely many solutions. For \(k = -3\), the system has no solution.

- Eliminate one variable while keeping \(a\) in the coefficients. - For which values of \(a\) does the resulting variable coefficient become zero? - Substitute each candidate value into the elimination equation. - Does elimination produce an identity or a contradiction?

1. Multiply equation (I) by \(4\): \(4ax - 36y = 24\). 2. Multiply equation (II) by \(a\): \(4ax - a^2y = 4a\). 3. Subtract the first transformed equation from the second: \((36 - a^2)y = 4a - 24\). 4. The system can fail to have exactly one solution only when \(36 - a^2 = 0\). Thus, \(a = 6\) or \(a = -6\). 5. If \(a = 6\), the elimination equation is \(0 = 0\), and the original equations are equivalent. The system has infinitely many solutions. 6. If \(a = -6\), the elimination equation is \(0 = -48\), a contradiction. The system has no solution.

The system has no solution for \(a = -6\) and infinitely many solutions for \(a = 6\).