Substitution/Elimination Method for Linear Equations + Explanations

To solve by substitution, solve one equation for one variable, then substitute into the other equation. From the first equation:

\(x + 2y = 10\)

Solve for \(x\):

\(x = 10 - 2y\)

Substitute \(x = 10 - 2y\) into the second equation:

\(2(10 - 2y) - y = 3\)

Expand and simplify:

\(20 - 4y - y = 3\)

\(20 - 5y = 3\)

\(-5y = -17\)

\(y = \frac{17}{5}\)

Substitute \(y = \frac{17}{5}\) back into \(x = 10 - 2y\):

\(x = 10 - 2\left(\frac{17}{5}\right)\)

\(x = \frac{50}{5} - \frac{34}{5} = \frac{16}{5}\)

The solution is \(\left(\frac{16}{5}, \frac{17}{5}\right)\).

To solve by elimination, add or subtract the equations to eliminate one variable. Start with:

\(2x + y = 7\) and \(3x - y = 8\)

Add the two equations together to eliminate \(y\):

\((2x + y) + (3x - y) = 7 + 8\)

\(5x = 15\)

Divide by 5 to solve for \(x\):

\(x = 3\)

Substitute \(x = 3\) into the first equation to solve for \(y\):

\(2(3) + y = 7\)

\(6 + y = 7\)

\(y = 1\)

The solution to the system is \((3, 1)\).

The formula for the midpoint of a line segment joining two points \((x_1, y_1)\) and \((x_2, y_2)\) is:

\(\text{Midpoint} = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\)

Substitute \((x_1, y_1) = (3, 4)\) and \((x_2, y_2) = (-1, 2)\):

\(\text{Midpoint} = \left(\frac{3 + (-1)}{2}, \frac{4 + 2}{2}\right)\)

\(\text{Midpoint} = \left(\frac{2}{2}, \frac{6}{2}\right)\)

\(\text{Midpoint} = (1, 3)\)

The midpoint is \((1, 3)\).

The distance formula between two points \((x_1, y_1)\) and \((x_2, y_2)\) is:

\(\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

Substitute \((x_1, y_1) = (1, 2)\) and \((x_2, y_2) = (4, 6)\):

\(\text{Distance} = \sqrt{(4 - 1)^2 + (6 - 2)^2}\)

\(\text{Distance} = \sqrt{3^2 + 4^2}\)

\(\text{Distance} = \sqrt{9 + 16}\)

\(\text{Distance} = \sqrt{25} = 5\)

The distance is \(5\).

To solve by elimination, add or subtract the equations to eliminate one variable. Start with:

\(5x + 3y = 16\) and \(2x - 3y = 4\)

Add the two equations to eliminate \(y\):

\((5x + 3y) + (2x - 3y) = 16 + 4\)

\(7x = 20\)

Solve for \(x\):

\(x = \frac{20}{7}\)

Substitute \(x = \frac{20}{7}\) into the first equation to solve for \(y\):

\(5\left(\frac{20}{7}\right) + 3y = 16\)

\(\frac{100}{7} + 3y = 16\)

\(3y = 16 - \frac{100}{7}\)

\(3y = \frac{112}{7} - \frac{100}{7}\)

\(3y = \frac{12}{7}\)

\(y = \frac{4}{7}\)

The solution is \(\left(\frac{20}{7}, \frac{4}{7}\right)\).

To solve by substitution, solve one equation for one variable, then substitute into the other equation. From the first equation:

\(3x - y = 7\)

Solve for \(y\):

\(y = 3x - 7\)

Substitute \(y = 3x - 7\) into the second equation:

\(2x + 3(3x - 7) = 1\)

Expand and simplify:

\(2x + 9x - 21 = 1\)

\(11x - 21 = 1\)

\(11x = 22\)

\(x = 2\)

Substitute \(x = 2\) into \(y = 3x - 7\):

\(y = 3(2) - 7\)

\(y = -1\)

The solution is \((2, -1)\).

The formula for the midpoint of a line segment joining two points \((x_1, y_1)\) and \((x_2, y_2)\) is:

\(\text{Midpoint} = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\)

Substitute \((x_1, y_1) = (-2, 5)\) and \((x_2, y_2) = (6, -3)\):

\(\text{Midpoint} = \left(\frac{-2 + 6}{2}, \frac{5 + (-3)}{2}\right)\)

\(\text{Midpoint} = \left(\frac{4}{2}, \frac{2}{2}\right)\)

\(\text{Midpoint} = (2, 1)\)

The midpoint is \((2, 1)\).

The distance formula between two points \((x_1, y_1)\) and \((x_2, y_2)\) is:

\(\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

Substitute \((x_1, y_1) = (0, 0)\) and \((x_2, y_2) = (3, 4)\):

\(\text{Distance} = \sqrt{(3 - 0)^2 + (4 - 0)^2}\)

\(\text{Distance} = \sqrt{3^2 + 4^2}\)

\(\text{Distance} = \sqrt{9 + 16}\)

\(\text{Distance} = \sqrt{25} = 5\)

The distance is \(5\).

To solve by elimination, add or subtract the equations to eliminate one variable. Start with:

\(4x - 5y = 2\) and \(6x + 5y = 18\)

Add the two equations to eliminate \(y\):

\((4x - 5y) + (6x + 5y) = 2 + 18\)

\(10x = 20\)

Solve for \(x\):

\(x = 2\)

Substitute \(x = 2\) into the first equation to solve for \(y\):

\(4(2) - 5y = 2\)

\(8 - 5y = 2\)

\(-5y = -6\)

\(y = \frac{6}{5}\)

The solution is \((2, \frac{6}{5})\).

To solve by substitution, substitute \(y = 2x - 1\) into the second equation:

\(3x + 2(2x - 1) = 13\)

Expand and simplify:

\(3x + 4x - 2 = 13\)

\(7x - 2 = 13\)

\(7x = 15\)

\(x = \frac{15}{7}\)

Substitute \(x = \frac{15}{7}\) into \(y = 2x - 1\):

\(y = 2\left(\frac{15}{7}\right) - 1\)

\(y = \frac{30}{7} - \frac{7}{7}\)

\(y = \frac{23}{7}\)

The solution is \(\left(\frac{15}{7}, \frac{23}{7}\right)\).

The formula for the midpoint of a line segment joining two points \((x_1, y_1)\) and \((x_2, y_2)\) is:

\(\text{Midpoint} = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\)

Substitute \((x_1, y_1) = (7, -2)\) and \((x_2, y_2) = (-5, 6)\):

\(\text{Midpoint} = \left(\frac{7 + (-5)}{2}, \frac{-2 + 6}{2}\right)\)

\(\text{Midpoint} = \left(\frac{2}{2}, \frac{4}{2}\right)\)

\(\text{Midpoint} = (1, 2)\)

The midpoint is \((1, 2)\).

The distance formula between two points \((x_1, y_1)\) and \((x_2, y_2)\) is:

\(\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

Substitute \((x_1, y_1) = (-3, 2)\) and \((x_2, y_2) = (4, -1)\):

\(\text{Distance} = \sqrt{(4 - (-3))^2 + (-1 - 2)^2}\)

\(\text{Distance} = \sqrt{(4 + 3)^2 + (-3)^2}\)

\(\text{Distance} = \sqrt{7^2 + 3^2}\)

\(\text{Distance} = \sqrt{49 + 9}\)

\(\text{Distance} = \sqrt{58}\)

The distance is \(\sqrt{58}\).

To solve by elimination, add or subtract the equations to eliminate one variable. Start with:

\(2x + 3y = 12\) and \(4x - 3y = 6\)

Add the two equations to eliminate \(y\):

\((2x + 3y) + (4x - 3y) = 12 + 6\)

\(6x = 18\)

Solve for \(x\):

\(x = 3\)

Substitute \(x = 3\) into the first equation to solve for \(y\):

\(2(3) + 3y = 12\)

\(6 + 3y = 12\)

\(3y = 6\)

\(y = 2\)

The solution is \((3, 2)\).

To solve by substitution, substitute \(y = x + 2\) into the second equation:

\(2x + (x + 2) = 10\)

Combine like terms:

\(3x + 2 = 10\)

Simplify:

\(3x = 8\)

\(x = \frac{8}{3}\)

Substitute \(x = \frac{8}{3}\) into \(y = x + 2\):

\(y = \frac{8}{3} + 2 = \frac{8}{3} + \frac{6}{3} = \frac{14}{3}\)

The solution is \(\left(\frac{8}{3}, \frac{14}{3}\right)\).

The formula for the midpoint of a line segment joining two points \((x_1, y_1)\) and \((x_2, y_2)\) is:

\(\text{Midpoint} = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\)

Substitute \((x_1, y_1) = (-3, 7)\) and \((x_2, y_2) = (5, -1)\):

\(\text{Midpoint} = \left(\frac{-3 + 5}{2}, \frac{7 + (-1)}{2}\right)\)

\(\text{Midpoint} = \left(\frac{2}{2}, \frac{6}{2}\right)\)

\(\text{Midpoint} = (1, 3)\)

The midpoint is \((1, 3)\).

The distance formula between two points \((x_1, y_1)\) and \((x_2, y_2)\) is:

\(\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

Substitute \((x_1, y_1) = (2, -5)\) and \((x_2, y_2) = (-4, 3)\):

\(\text{Distance} = \sqrt{(-4 - 2)^2 + (3 - (-5))^2}\)

\(\text{Distance} = \sqrt{(-6)^2 + (8)^2}\)

\(\text{Distance} = \sqrt{36 + 64}\)

\(\text{Distance} = \sqrt{100} = 10\)

The distance is \(10\).