Systems of Equations by Elimination, SAT math lesson

Two shopping receipts. On the first, 3 pens and 2 notebooks cost 12 dollars. On the second, 3 pens and 4 notebooks cost 18 dollars. Lay one receipt under the other. The pens are the same, so the second receipt is 2 notebooks and 6 dollars more, and a notebook costs 3 dollars. Back on the first receipt, 2 notebooks cost 6 dollars, which leaves 6 dollars for 3 pens, so a pen costs 2 dollars. Nobody had to solve for anything first. Elimination is that move made systematic, combining whole equations so that one variable drops out.

See also Linear Systems in Two and Three Variables in Precalculus.

Key terms

Linear combination of equations

Definition Given two equations $L_1 = R_1$ and $L_2 = R_2$ and real numbers $p$ and $q$ , their linear combination is the equation

pL_1 + qL_2 = pR_1 + qR_2.

Each equation is multiplied through, left side and right side, and then the results are added. If a pair $(x, y)$ satisfies both original equations, it also satisfies every linear combination of them. Taking $q = -1$ subtracts the second equation.

In plain words Scale each equation by a number, then add them side by side. Equal things added to equal things stay equal.

Example Take $x + 2y = 5$ and $3x - y = 1$ with $p = 1$ and $q = 2$ . The second equation times $2$ is $6x - 2y = 2$ , with the right side doubled too. Adding gives

7x + 0y = 7, \quad \text{that is} \quad x = 1.

Forgetting to double the right side would give $6x - 2y = 1$ and then the wrong value $x = \frac{6}{7}$ .

Equivalent system

Definition Two systems are equivalent when they have the same solution set. A system stays equivalent when you

multiply one equation by a nonzero number, or
replace one equation by its sum with a multiple of the other equation.

Both steps can be undone. For step 2, if the new second equation is $E_2 + cE_1$ and $E_1$ is kept, then subtracting $c$ times $E_1$ gives back $E_2$ . So every solution of the new system solves the old one and the other way around. Keep one equation as it is while you replace the other. Replacing both equations by multiples of the same single equation would lose information.

In plain words Elimination never changes the answer, because each step can be reversed. You just trade the system for an easier one with the same solutions.

Example The systems

\begin{aligned} x + 2y &= 5 \\[0.4em] 3x - y &= 1 \end{aligned} \qquad \text{and} \qquad \begin{aligned} x + 2y &= 5 \\[0.4em] 7x &= 7 \end{aligned}

are equivalent, since the second equation on the right is the first equation plus $2$ times the second one on the left. The right system is easy. From $7x = 7$ comes $x = 1$ , and then $1 + 2y = 5$ gives $y = 2$ . Both systems have the single solution $(1, 2)$ .

Elimination

Definition The elimination method solves a system of two linear equations by forming a linear combination in which the coefficient of one variable is $0$ . Solve the resulting one-variable equation, then substitute that value into an original equation to find the other variable, and check the pair in both original equations. When the coefficients of one variable are already opposites, simply add the equations. When they are equal, subtract.

In plain words Line up the $x$ terms, the $y$ terms and the constants in columns. If one column already cancels, add and you are left with a single unknown.

Example Solve $x + y = 10$ and $x - y = 4$ . The $y$ terms are opposites, so add.

\begin{aligned} 2x &= 14 \\[0.4em] x &= 7 \end{aligned}

Substituting into $x + y = 10$ gives $y = 3$ . The pair $(7, 3)$ checks in both, since $7 + 3 = 10$ and $7 - 3 = 4$ .

Multipliers on both equations

Definition If the coefficients of a variable are $a_1$ and $a_2$ , both nonzero, multiplying the first equation by $a_2$ and the second by $-a_1$ makes that variable cancel when you add. Any other pair of multipliers that makes the two coefficients opposites works as well. With integer coefficients a smaller choice is often possible, using the least common multiple of their absolute values.

In plain words Make the two coefficients of one variable equal in size and opposite in sign. Usually that means multiplying each equation by the other one’s coefficient.

Example Solve $3x + 2y = 10$ and $5x - 3y = 23$ . The $y$ coefficients are $2$ and $-3$ , so multiply the first equation by $3$ and the second by $2$ .

\begin{aligned} 9x + 6y &= 30 \\[0.4em] 10x - 6y &= 46 \end{aligned}

Adding gives $19x = 76$ , so $x = 4$ . Then $12 + 2y = 10$ gives $y = -1$ . In the second original equation, $20 + 3 = 23$ , so the solution is $(4, -1)$ .

Substitution or elimination

Definition Substitution solves one equation for a variable and replaces that variable in the other equation. Both methods produce equivalent systems, so they always reach the same solution set. Substitution is usually faster when a variable is already isolated or has coefficient $1$ or $-1$ . Elimination is usually faster when both equations are in the form $ax + by = c$ with no coefficient of $\pm 1$ , since isolating a variable there would create fractions.

In plain words Use the method that keeps the numbers whole. An equation that starts with $y =$ is asking for substitution.

Example For $y = 4x - 3$ and $2x + 3y = 19$ , substitution is natural. The second equation becomes $2x + 12x - 9 = 19$ , so $14x = 28$ and $x = 2$ , which gives $y = 5$ . For $2x + 3y = 7$ and $3x + 4y = 10$ , elimination is easier. Multiplying by $3$ and $2$ gives $6x + 9y = 21$ and $6x + 8y = 20$ , and subtracting leaves $y = 1$ , so $x = 2$ .

Solution check

Definition A pair $(x, y)$ is a solution of the system exactly when it makes each original equation true. To check, substitute the pair into both original equations, not into the scaled versions. A scaled equation copied with a mistake can agree with a wrong answer, while the originals cannot.

In plain words Plug your answer back into both equations you were given. If either one fails, something went wrong.

Example Suppose an arithmetic slip while solving $3x + 2y = 10$ and $5x - 3y = 23$ leads to the candidate $(2, 2)$ . The first equation passes, since $6 + 4 = 10$ . The second fails, since $10 - 6 = 4 \ne 23$ . Checking both is what exposes the error.

Target linear combination

Definition Some questions ask for the value of an expression such as $x + y$ or $4x + 6y$ rather than for $x$ and $y$ . If numbers $p$ and $q$ make the left side of $pE_1 + qE_2$ equal to the target expression, then the right side of that combination is its value. You can find $p$ and $q$ by matching coefficients, though often adding or subtracting the equations is enough.

In plain words Look at what the question wants before solving. Adding or subtracting the equations often produces it in one step.

Example If $3x + 5y = 22$ and $x + 3y = 10$ , what is $x + y$ ? Subtracting the second equation from the first gives

2x + 2y = 12,

so $x + y = 6$ . Solving fully confirms it, since the solution is $(4, 2)$ .

Check yourself

Which ordered pair $(x, y)$ is the solution of the system $4x + 3y = 5$ and $6x - 2y = 27$ ? (A) $\left(\frac{7}{2}, -3\right)$ (B) $\left(-3, \frac{7}{2}\right)$ (C) $(2, -1)$ (D) $\left(4, -\frac{3}{2}\right)$

Show the answer

The answer is (A). Multiply the first equation by $2$ and the second by $3$ .

\begin{aligned} 8x + 6y &= 10 \\[0.4em] 18x - 6y &= 81 \end{aligned}

Adding gives $26x = 91$ , so $x = \frac{7}{2}$ , and then $14 + 3y = 5$ gives $y = -3$ . Choice (B) swaps the coordinates. Choice (C) satisfies only the first equation, and choice (D) satisfies only the second.

If $5x + 2y = 31$ and $3x + 4y = 27$ , what is the value of $x - y$ ?

Show the answer

The answer is $2$ . Subtracting the second equation from the first gives $2x - 2y = 4$ , so $x - y = 2$ . The full solution $(5, 3)$ confirms it.

Which system has the same solution set as the system $x + 2y = 5$ and $3x - y = 1$ ? (A) $x + 2y = 5$ and $7x = 7$ (B) $x + 2y = 5$ and $6x - 2y = 1$ (C) $x + 2y = 5$ and $7x = 11$ (D) $3x - y = 1$ and $7x = 6$

Show the answer

The answer is (A). Adding $2$ times the second equation to the first gives $7x = 7$ , and keeping $x + 2y = 5$ makes the new system equivalent. Both systems have the solution $(1, 2)$ . Choice (B) multiplies only the left side of $3x - y = 1$ by $2$ . Choices (C) and (D) come from adding with a wrong right side. Choice (C) doubles the $5$ instead of the $1$ , since $10 + 1 = 11$ , and choice (D) doubles neither, since $5 + 1 = 6$ . They give $x = \frac{11}{7}$ and $x = \frac{6}{7}$ instead of $x = 1$ .

Systems of Equations by Elimination

Theory, part 31

Practice, part 32

Key terms

Linear combination of equations

Equivalent system

Elimination

Multipliers on both equations

Substitution or elimination

Solution check

Target linear combination

Check yourself

How this lesson connects

Builds on

Used later in