Grade 9 Linear Equations - Substitution, Elimination Methods Questions

Lines that are parallel have the same slope. The given equation is in slope-intercept form \(y = mx + b\), where \(m\) is the slope. The slope of the given line is \(-5\), so the slope of any parallel line is also \(-5\).

Use the point-slope form \(y - y_1 = m(x - x_1)\), where \(m = 3\) and \((x_1, y_1) = (2, -1)\). Substituting: \(y + 1 = 3(x - 2)\). Simplify: \(y + 1 = 3x - 6\). Subtract 1: \(y = 3x - 7\).

To find the slope, rewrite the equation in slope-intercept form \(y = mx + b\). Start by isolating \(y\): \(-2y = -4x + 8\). Divide by \(-2\): \(y = 2x - 4\). The slope is \(2\).

The \(x\)-intercept occurs when \(y = 0\). Substitute \(y = 0\) into the equation: \(0 = -2x + 6\). Solving for \(x\): \(2x = 6\), so \(x = 3\). Therefore, the \(x\)-intercept is \(3\).

Two lines are perpendicular if the product of their slopes is \(-1\). The slopes of the lines are \(4\) and \(-\frac{1}{4}\). Their product is \(4 \times -\frac{1}{4} = -1\), so the lines are perpendicular.

First, calculate the slope \(m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{8 - (-1)}{5 - 2} = \frac{9}{3} = 3\). Using point-slope form \(y - y_1 = m(x - x_1)\) with (2, -1): \(y - (-1) = 3(x - 2)\). Simplify: \(y + 1 = 3x - 6\), and subtract 1: \(y = 3x - 7\).

To find the \(x\)-intercept, set \(y = 0\): \(3x = 12\), so \(x = 4\). The \(x\)-intercept is \((4, 0)\). To find the \(y\)-intercept, set \(x = 0\): \(4y = 12\), so \(y = 3\). The \(y\)-intercept is \((0, 3)\).

Set \(C(x) = 450\): \(5x + 200 = 450\). Subtract 200: \(5x = 250\). Divide by 5: \(x = 50\). Therefore, 50 items must be produced to achieve the total cost of $450.

Set the equations equal to find the \(x\)-coordinate: \(\frac{1}{2}x + 3 = -2x - 5\). Multiply through by 2: \(x + 6 = -4x - 10\). Add \(4x\): \(5x + 6 = -10\). Subtract 6: \(5x = -16\). Divide by 5: \(x = -\frac{16}{5}\). Substitute into \(y = \frac{1}{2}x + 3\): \(y = \frac{1}{2}(-\frac{16}{5}) + 3 = -\frac{8}{5} + 3 = \frac{7}{5}\). The intersection point is \((-\frac{16}{5}, \frac{7}{5})\).

The formula for perimeter is \(P = 2L + 2W\). Substitute \(L = 2x + 4\) and \(W = x - 2\): \(64 = 2(2x + 4) + 2(x - 2)\). Simplify: \(64 = 4x + 8 + 2x - 4\). Combine like terms: \(64 = 6x + 4\). Subtract 4: \(60 = 6x\). Divide by 6: \(x = 10\).

When a line is reflected over the \(y\)-axis, the \(x\)-coordinate is negated. This affects the slope, which is the coefficient of \(x\). For \(y = 2x - 1\), the slope becomes \(-2\), so the equation of the reflected line is \(y = -2x - 1\).

Perpendicular lines have slopes that are negative reciprocals. The slope of the given line is \(\frac{3}{4}\), so the perpendicular slope is \(-\frac{4}{3}\). Using the point-slope form with slope \(-\frac{4}{3}\) and point (-4, 1): \(y - 1 = -\frac{4}{3}(x + 4)\). Expand: \(y - 1 = -\frac{4}{3}x - \frac{16}{3}\). Add 1: \(y = -\frac{4}{3}x - \frac{13}{3}\).

The slope is calculated as \(m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{9 - 3}{-2 - 1} = \frac{6}{-3} = -2\). Since the slope is negative, the line is decreasing.

The slope of the first line is \(4\), so the perpendicular slope is \(-\frac{1}{4}\). Use point-slope form with slope \(-\frac{1}{4}\) and point (2, -3): \(y + 3 = -\frac{1}{4}(x - 2)\). Expand: \(y + 3 = -\frac{1}{4}x + \frac{1}{2}\). Subtract 3: \(y = -\frac{1}{4}x - \frac{5}{2}\).

Find the \(x\)-intercept by setting \(y = 0\): \(2x = 20\), so \(x = 10\). Find the \(y\)-intercept by setting \(x = 0\): \(5y = 20\), so \(y = 4\). The sum of the intercepts is \(10 + 4 = 14\), not \(10\). The given statement is incorrect.

Substitute \(y = 2x + 3\) into \(3x - y = 4\): \(3x - (2x + 3) = 4\). Simplify: \(x - 3 = 4\), so \(x = 7\). Substitute \(x = 7\) back into \(y = 2x + 3\): \(y = 2(7) + 3 = 17\). The solution is \((7, 17)\).

Add the two equations: \((2x + y) + (3x - y) = 7 + 8\). This simplifies to \(5x = 15\), so \(x = 3\). Substitute \(x = 3\) into \(2x + y = 7\): \(2(3) + y = 7\), so \(y = 1\). The solution is \((3, 1)\).

The first equation \(y = -x + 4\) has a slope of \(-1\) and y-intercept \(4\). The second equation \(y = 2x - 1\) has a slope of \(2\) and y-intercept \(-1\). Plotting both lines, they intersect at \((1, 1)\). Verify by substitution: \(y = -1 + 4 = 1\) and \(y = 2(1) - 1 = 1\).

Multiply the second equation by \(2\) to align the coefficients of \(x\): \(4x + 6y = 28\). Subtract the first equation: \((4x + 6y) - (4x + 5y) = 28 - 20\), so \(y = 8\). Substitute \(y = 8\) into \(4x + 5y = 20\): \(4x + 5(8) = 20\), so \(4x = -20\) and \(x = -5\). The solution is \((-5, 8)\).

Rewriting \(4x - 2y = 8\) as \(2(2x - y) = 8\) shows it is a multiple of \(2x - y = 4\). The two equations are identical, representing the same line. Therefore, there are infinitely many solutions.

Substitute \(y = 2x - 4\) into \(3x + y = 7\): \(3x + (2x - 4) = 7\). Combine like terms: \(5x - 4 = 7\), so \(5x = 11\) and \(x = \frac{11}{5}\). Substitute \(x = \frac{11}{5}\) back into \(y = 2x - 4\): \(y = 2(\frac{11}{5}) - 4 = \frac{22}{5} - \frac{20}{5} = \frac{2}{5}\). The solution is \((\frac{11}{5}, \frac{2}{5})\).

Multiply the second equation by 2 to match coefficients: \(6x + 4y = 20\) and \(6x + 4y = 20\). Since the equations are identical, they represent the same line, and there are infinitely many solutions.

Rearrange \(x + y = 5\) to \(x = 5 - y\). Substitute into \(2x - 3y = 6\): \(2(5 - y) - 3y = 6\). Expand: \(10 - 2y - 3y = 6\). Combine terms: \(-5y = -4\), so \(y = \frac{4}{5}\). Substitute \(y = \frac{4}{5}\) into \(x = 5 - y\): \(x = 5 - \frac{4}{5} = \frac{25}{5} - \frac{4}{5} = \frac{21}{5}\). The solution is \((\frac{21}{5}, \frac{4}{5})\).

Rewriting \(4x - 2y = -2\): \(-2y = -4x - 2\), so \(y = 2x + 1\). The equations are identical, meaning the system represents the same line. Thus, there are infinitely many solutions.

Add the equations: \((5x + 2y) + (3x - 2y) = 18 + 2\). This simplifies to \(8x = 20\), so \(x = \frac{20}{8} = 2.5\). Substitute \(x = 2.5\) into \(5x + 2y = 18\): \(5(2.5) + 2y = 18\), so \(12.5 + 2y = 18\). Simplify: \(2y = 5.5\), so \(y = 2.75\). The solution is \((2.5, 2.75)\).

Multiply the first equation by 2: \(8x + 2y = 18\). Subtract the second equation: \((8x + 2y) - (3x + 2y) = 18 - 11\), so \(5x = 7\) and \(x = \frac{7}{5}\). Substitute \(x = \frac{7}{5}\) into \(4x + y = 9\): \(4(\frac{7}{5}) + y = 9\), so \(\frac{28}{5} + y = 9\). Simplify: \(y = 9 - \frac{28}{5} = \frac{45}{5} - \frac{28}{5} = \frac{17}{5}\). The solution is \((\frac{7}{5}, \frac{17}{5})\).

Substitute \(y = 4x - 7\) into \(2x + y = 5\): \(2x + (4x - 7) = 5\). Combine terms: \(6x - 7 = 5\), so \(6x = 12\) and \(x = 2\). Substitute \(x = 2\) into \(y = 4x - 7\): \(y = 4(2) - 7 = 8 - 7 = 1\). The solution is \((2, 1)\).

Multiply the second equation by 3 to match coefficients of \(y\): \(3(4x - y) = 3(6)\), giving \(12x - 3y = 18\). Add this to the first equation: \((2x + 3y) + (12x - 3y) = 12 + 18\). Simplify: \(14x = 30\), so \(x = \frac{15}{7}\). Substitute \(x = \frac{15}{7}\) into \(4x - y = 6\): \(4(\frac{15}{7}) - y = 6\). Simplify: \(\frac{60}{7} - y = 6\), so \(y = \frac{60}{7} - \frac{42}{7} = \frac{18}{7}\). The solution is \((\frac{15}{7}, \frac{18}{7})\).

Substitute \(y = x + 5\) into \(2x + 3y = 16\): \(2x + 3(x + 5) = 16\). Expand: \(2x + 3x + 15 = 16\). Combine terms: \(5x + 15 = 16\), so \(5x = 1\) and \(x = \frac{1}{5}\). Substitute \(x = \frac{1}{5}\) into \(y = x + 5\): \(y = \frac{1}{5} + 5 = \frac{1}{5} + \frac{25}{5} = \frac{26}{5}\). The solution is \((\frac{1}{5}, \frac{26}{5})\).

Multiply the first equation by 2 to match coefficients: \(2(3x + 4y) = 2(8)\), giving \(6x + 8y = 16\). Add the second equation: \((6x + 8y) + (6x - 8y) = 16 - 4\). Simplify: \(12x = 12\), so \(x = 1\). Substitute \(x = 1\) into \(3x + 4y = 8\): \(3(1) + 4y = 8\), so \(4y = 5\) and \(y = \frac{5}{4}\). The solution is \((1, \frac{5}{4})\).

Rearrange \(x - y = 3\) to \(x = y + 3\). Substitute into \(x + y = 7\): \((y + 3) + y = 7\). Combine terms: \(2y + 3 = 7\), so \(2y = 4\) and \(y = 2\). Substitute \(y = 2\) into \(x = y + 3\): \(x = 2 + 3 = 5\). The solution is \((5, 2)\).

Multiply the second equation by 3: \(3(4x - y) = 3(10)\), giving \(12x - 3y = 30\). Add to the first equation: \((2x + 3y) + (12x - 3y) = 14 + 30\). Simplify: \(14x = 44\), so \(x = \frac{22}{7}\). Substitute \(x = \frac{22}{7}\) into \(4x - y = 10\): \(4(\frac{22}{7}) - y = 10\), so \(\frac{88}{7} - y = 10\). Simplify: \(y = \frac{88}{7} - \frac{70}{7} = \frac{18}{7}\). The solution is \((\frac{22}{7}, \frac{18}{7})\).

Rearrange \(x + 2y = 10\) to \(x = 10 - 2y\). Substitute into \(2x - 3y = -1\): \(2(10 - 2y) - 3y = -1\). Expand: \(20 - 4y - 3y = -1\). Combine terms: \(20 - 7y = -1\), so \(-7y = -21\) and \(y = 3\). Substitute \(y = 3\) into \(x = 10 - 2y\): \(x = 10 - 6 = 4\). The solution is \((4, 3)\).

Multiply the first equation by 3: \(3(5x - y) = 3(7)\), giving \(15x - 3y = 21\). Add to the second equation: \((15x - 3y) + (10x + 3y) = 21 + 1\). Simplify: \(25x = 22\), so \(x = \frac{22}{25}\). Substitute \(x = \frac{22}{25}\) into \(5x - y = 7\): \(5(\frac{22}{25}) - y = 7\). Simplify: \(\frac{110}{25} - y = 7\), so \(y = \frac{110}{25} - \frac{175}{25} = -\frac{65}{25}\). The solution is \((\frac{22}{25}, -\frac{65}{25})\).