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To verify an ordered pair solves a linear equation with fractions, substitute the $x$ and $y$ values into the left-hand side of the equation. Evaluate the products by multiplying each fraction by its corresponding integer.

Conditions: The candidate solution values must already be known.; The equation being checked contains those variables with fractional coefficients.

To verify an ordered pair solves a linear equation with fractions, substitute the $x$ and $y$ values into the left-hand side of the equation. Evaluate the products by multiplying each fraction by its corresponding integer.

Conditions: The candidate solution values must already be known.; The equation being checked contains those variables with fractional coefficients.

The two equations can be added directly to eliminate $y$ because their $y$-coefficients are exact opposites ($-4y$ and $+4y$). When equations with opposite coefficients for the same variable are added term by term, those terms cancel out to zero, leaving a one-variable equation in $x$.

Conditions: The system has been reduced to integer coefficients: $3x - 4y = 14$ and $15x + 4y = 22$.; The equations are aligned term-by-term before addition.

The two equations can be added directly to eliminate $y$ because their $y$-coefficients are exact opposites ($-4y$ and $+4y$). When equations with opposite coefficients for the same variable are added term by term, those terms cancel out to zero, leaving a one-variable equation in $x$.

Conditions: The system has been reduced to integer coefficients: $3x - 4y = 14$ and $15x + 4y = 22$.; The equations are aligned term-by-term before addition.

The solution is checked in the other original equation to confirm that the ordered pair satisfies the entire system, not just the single equation used for back-substitution. If the pair was found by substituting into the first equation, checking it in the second equation verifies that both equations are simultaneously true for those values.

Conditions: A candidate ordered pair has been found.; At least one original equation remains available for substitution checking.; The equation used for back-substitution is not reused for the check.

The solution is checked in the other original equation to confirm that the ordered pair satisfies the entire system, not just the single equation used for back-substitution. If the pair was found by substituting into the first equation, checking it in the second equation verifies that both equations are simultaneously true for those values.

Conditions: A candidate ordered pair has been found.; At least one original equation remains available for substitution checking.; The equation used for back-substitution is not reused for the check.

The fraction $\frac{22}{20}$ simplifies to $\frac{11}{10}$ by dividing both the numerator and the denominator by their greatest common factor, which is 2. Since $22 \div 2 = 11$ and $20 \div 2 = 10$, the reduced fraction is $\frac{11}{10}$.

Conditions: The numerator (22) and denominator (20) share a common factor greater than 1.; The same nonzero factor (2) divides both the numerator and the denominator.

The fraction $\frac{22}{20}$ simplifies to $\frac{11}{10}$ by dividing both the numerator and the denominator by their greatest common factor, which is 2. Since $22 \div 2 = 11$ and $20 \div 2 = 10$, the reduced fraction is $\frac{11}{10}$.

Conditions: The numerator (22) and denominator (20) share a common factor greater than 1.; The same nonzero factor (2) divides both the numerator and the denominator.

The equation is multiplied by 3 to clear the denominators and eliminate the fractions. Since the denominators present are 1 and 3, the common denominator is 3.

Conditions: The equation contains fractional terms with denominators 1 and 3.; All terms on both sides of the equation are multiplied by the same nonzero number (3).

The equation is multiplied by 3 to clear the denominators and eliminate the fractions. Since the denominators present are 1 and 3, the common denominator is 3.

Conditions: The equation contains fractional terms with denominators 1 and 3.; All terms on both sides of the equation are multiplied by the same nonzero number (3).

A common denominator is required before subtracting $\frac{6}{4}$ and $\frac{2}{5}$ because the fractions have different denominators. The rule for subtracting fractions states that $\frac{a}{c} - \frac{b}{c} = \frac{a-b}{c}$, which only applies when both fractions share the same denominator $c$.

Conditions: The fractions $\frac{6}{4}$ and $\frac{2}{5}$ have unlike denominators (4 and 5).; The operation being performed is subtraction.

A common denominator is required before subtracting $\frac{6}{4}$ and $\frac{2}{5}$ because the fractions have different denominators. The rule for subtracting fractions states that $\frac{a}{c} - \frac{b}{c} = \frac{a-b}{c}$, which only applies when both fractions share the same denominator $c$.

Conditions: The fractions $\frac{6}{4}$ and $\frac{2}{5}$ have unlike denominators (4 and 5).; The operation being performed is subtraction.

Multiplying a fraction by $\frac{5}{5}$ or $\frac{4}{4}$ keeps it equivalent because each of these factors equals 1. Any nonzero number divided by itself equals 1, and multiplying any quantity by 1 does not change its value.

Conditions: The factor used is a fraction whose numerator and denominator are equal and nonzero.; The original fraction is being rewritten to obtain a specific common denominator.

Multiplying a fraction by $\frac{5}{5}$ or $\frac{4}{4}$ keeps it equivalent because each of these factors equals 1. Any nonzero number divided by itself equals 1, and multiplying any quantity by 1 does not change its value.

Conditions: The factor used is a fraction whose numerator and denominator are equal and nonzero.; The original fraction is being rewritten to obtain a specific common denominator.