Solving Systems of Linear Equations

To solve a system by graphing, draw both lines on one coordinate plane and look for a point they share. In the figure below, the horizontal axis is the $x$-axis and the vertical axis is the $y$-axis.

Slope is the change in $y$ over the change in $x$, for a nonzero change in $x$. A horizontal line has slope $0$, and a vertical line has undefined slope.

Take the system $y = 2x - 1$ and $y = -x + 5$. Line A is $y = 2x - 1$, with slope $2$, and it passes through Point A at $(0, -1)$. Line B is $y = -x + 5$, with slope $-1$, and it passes through Point B at $(0, 5)$. The lines appear to cross at Point C, which looks like $(2, 3)$. A drawing only suggests this. It cannot prove exact coordinates. So call $(2, 3)$ a candidate: a possible solution that you must test in both equations.

Two lines do not always meet at exactly one point. If two equations describe the same line, every point on it is a solution, so the system has infinitely many solutions. If the lines are different and parallel, the system has no solution.

Parallel lines are lines in the same plane that never meet. The Parallel Slopes Theorem says that two different nonvertical lines are parallel exactly when their slopes are equal. It works in both directions: equal slopes give parallel lines, and parallel lines have equal slopes. So different slopes mean the lines are not parallel, and they must meet.

The next figure, with the same axes, shows $y = 2x + 1$ as Line A, through Point A at $(0, 1)$, and $y = 2x - 3$ as Line B, through Point B at $(0, -3)$. Each equation has the form $y = mx + b$, so both lines are nonvertical. Both slopes are $2$ and the $y$-intercepts differ, so these are two different lines with equal slopes. By the direction "equal slopes give parallel lines," they are parallel. That reason comes from the equations, not from how the picture looks, and it means this system has no solution.

A graph shows what a solution means, but it cannot always give exact values. The lines $y = 2x - 1$ and $y = -x + 4$ meet at $\left(\frac{5}{3}, \frac{7}{3}\right)$, since both equations give $y = \frac{7}{3}$ when $x = \frac{5}{3}$. No one can read those exact values from a drawing. Algebra can find them.

Substitution rests on the substitution property of equality: if two quantities are equal, you may replace one with the other in any equation. At a solution, $x$ and $y$ stand for the same numbers in both equations. So if one equation says $y = x - 1$, you may write $x - 1$ in place of $y$ in the other equation. That leaves one equation in one variable, which you already know how to solve.

The plan: get one variable alone, replace it in the other equation, solve, find the second variable, then check the pair in both original equations. If getting a variable alone means dividing by its coefficient, the number that multiplies it, that number must be nonzero. Each step follows from the one before, so any solution must have the values you find. The check confirms that those values work.

Elimination removes one variable by adding the two equations. It rests on the addition property of equality: if $a = b$ and $c = d$, then $a + c = b + d$. At a solution, both equations are true, so the sum of the left sides equals the sum of the right sides.

When a variable's coefficients in the two equations are opposites, such as $2$ and $-2$, their sum is $0$, so adding the equations makes that variable disappear.

If no coefficients are opposites yet, multiply both sides of one equation, or of each one, by a nonzero number, using the multiplication property of equality. The number must be nonzero, because multiplying by $0$ turns any equation into $0 = 0$ and erases its line. A nonzero multiplier keeps the same line.

Sometimes both variables disappear as you use substitution or elimination. The statement left over tells you which case you have.

Try substitution on $y = 2x + 1$ and $y = 2x - 3$. By the substitution property of equality, replacing $y$ gives $2x + 1 = 2x - 3$. Subtracting $2x$ from both sides, by the subtraction property of equality, leaves $1 = -3$. Any solution would have to make this statement true. It is false, so the system has no solution, which agrees with the parallel lines above.

Now try $y = x + 1$ and $2y = 2x + 2$. Replacing $y$ the same way gives $2(x + 1) = 2x + 2$, and the distributive property makes this $2x + 2 = 2x + 2$. Subtracting $2x + 2$ from both sides leaves $0 = 0$, which is true for every $x$. Dividing $2y = 2x + 2$ by the nonzero number $2$ gives $y = x + 1$, so both equations describe the same line, and every point on it is a solution.

To compare slopes in the table below, solve each equation for $y$. That divides by the coefficient of $y$, which must be nonzero. If that coefficient is $0$, the line is vertical, like $x = 3$.

All three methods look for the same points, so you may choose the one that fits the form of the equations.

• Graphing shows how the lines relate, but it gives only a candidate to test.
• Substitution is quick when a variable is alone or has a coefficient of $1$ or $-1$.
• Elimination is quick when a variable's coefficients are opposites or can become opposites with a nonzero multiplier.

These are guides, not rules. In the table below, test each listed solution in both of its equations.

A solution of a system is a point on both lines, so it must make both equations true. Graphing shows where to look, and substitution and elimination give exact values. Choose the method that fits the form of the equations, then test your answer in both of them.