6-4 Elimination Using Multiplication

Property

If adding or subtracting the equations directly does not eliminate a variable, you must alter them first. When one coefficient is a simple multiple of the other, you only need to multiply ONE equation by a constant.

By multiplying every single term on both sides of that equation by the chosen constant, you create an equivalent equation with the exact opposite coefficient needed for elimination.

Examples

• Example 1 (Multiplying One Equation): Solve the system $3x + y = 5$ and $2x - 3y = 7$.

The $y$ coefficients are $1$ and $-3$. To make them opposites, multiply the ENTIRE top equation by $3$:
$3(3x + y) = 3(5) \rightarrow 9x + 3y = 15$
Now, add this new equation to the bottom equation:
$(9x + 2x) + (3y - 3y) = 15 + 7$
$11x = 22 \rightarrow x = 2$

• Example 2 (Back-Substitution): Now that $x = 2$, substitute it back into the original, simplest equation ($3x + y = 5$):

$3(2) + y = 5 \rightarrow 6 + y = 5 \rightarrow y = -1$.
The solution is $(2, -1)$.

Explanation

You can think of an equation like a recipe. If a recipe makes 1 batch of cookies, multiplying every single ingredient by 3 gives you 3 batches, but it is still the exact same recipe! In algebra, multiplying the left and right sides by the same number keeps the line exactly the same, but it changes the numbers to fit your needs. The most common mistake is multiplying the variables but forgetting to multiply the constant on the other side of the equal sign.

Property

When the coefficients of the variable you want to eliminate are not simple multiples of each other, you must multiply BOTH equations by different constants.

Find the Least Common Multiple (LCM) of the two coefficients. Then, multiply each equation by a constant that turns that variable's coefficient into the LCM, ensuring one is positive and one is negative so they sum to zero.

Examples

• Example 1 (Eliminate y): Solve $3x + 4y = 10$ and $5x + 6y = 14$.

The LCM of the $y$ coefficients (4 and 6) is 12.
Multiply the first equation by $3$ to get $12y$, and the second by $-2$ to get $-12y$:
$3(3x + 4y = 10) \rightarrow 9x + 12y = 30$
$-2(5x + 6y = 14) \rightarrow -10x - 12y = -28$
Add the new equations: $(9x - 10x) + (12y - 12y) = 30 - 28 \rightarrow -x = 2 \rightarrow x = -2$.

• Example 2 (Back-Substitution): Substitute $x = -2$ into the first original equation:

$3(-2) + 4y = 10 \rightarrow -6 + 4y = 10 \rightarrow 4y = 16 \rightarrow y = 4$.
The solution is $(-2, 4)$.

Explanation

When neither number easily scales into the other, you have to find a common meeting point. This is exactly like finding a common denominator when adding fractions! If you have a 4 and a 6, they both comfortably meet at 12. You scale the top equation up to reach 12, and you scale the bottom equation to reach -12. Once they are perfect opposites, you add them together to trigger the elimination.

Property

When you scale equations using multiplication and add them together, it is possible for both variables to cancel out simultaneously.

• Contradiction: If the result is a false statement (e.g., $0 = c$, where $c \neq 0$), the lines are parallel, resulting in No Solution.
• Identity: If the result is a true statement (e.g., $0 = 0$), the equations represent the exact same line, resulting in Infinitely Many Solutions.

Examples

• No Solution: Solve $2x + 3y = 6$ and $4x + 6y = 15$.

Multiply the top equation by $-2$ to eliminate $x$:
$-2(2x + 3y = 6) \rightarrow -4x - 6y = -12$
Add this to the second equation:
$(-4x + 4x) + (-6y + 6y) = -12 + 15 \rightarrow 0 = 3$.
This is a false statement, so there is no solution.

• Infinitely Many Solutions: Solve $3x - y = 4$ and $6x - 2y = 8$.

Multiply the top equation by $-2$:
$-2(3x - y = 4) \rightarrow -6x + 2y = -8$
Add this to the second equation:
$(-6x + 6x) + (2y - 2y) = -8 + 8 \rightarrow 0 = 0$.
This is a true statement, so there are infinitely many solutions.

Explanation

When both variables vanish, your algebra is uncovering a secret about the two equations. If you multiply an equation and it suddenly looks identical to the other one (giving $0=0$), it means they were the exact same line the entire time—just disguised by a scale factor! If multiplying makes the left sides match but the right sides completely different (giving $0=3$), it means they share a slope but have different intercepts, proving they are parallel lines that will never cross.