4 Algebra Flashback \#12 11. Use the elimination method by subtraction to solve for \( x \) and then \( y \). Equation 1 Equation 2

To solve the system of equations using the elimination method, you want to manipulate the equations so that you can eliminate one variable. Suppose you have two equations: Equation 1: \( 2x + 3y = 6 \) Equation 2: \( 4x - 5y = 7 \) First, you can multiply Equation 1 by 2 to align the coefficients of \( x \): \( 4x + 6y = 12 \) Now, you can subtract Equation 2 from this new equation: \( (4x + 6y) - (4x - 5y) = 12 - 7 \) This simplifies to: \( 6y + 5y = 5 \) Thus, \( 11y = 5 \) From this, you find \( y = \frac{5}{11} \). Next, substitute \( y \) back into either original equation to find \( x \). Using Equation 1: \( 2x + 3\left(\frac{5}{11}\right) = 6 \) This leads you to solve for \( x \): \( 2x + \frac{15}{11} = 6 \) Multiply the entire equation by 11 to eliminate the fraction: \( 22x + 15 = 66 \) Thus, \( 22x = 51 \) or \( x = \frac{51}{22} \). So the final solution is \( x = \frac{51}{22} \) and \( y = \frac{5}{11} \). The elimination method shines in solving systems of equations because it allows you to reduce complex problems to their simplest forms. Just imagine two competitors racing to the finish line, but instead, they are variables working together to eliminate each other until only one solution remains! It’s like a mathematical game of hide and seek where the variables are trying to play tricks, but in the end, they’re found and pinned down to their numeric values. If you find yourself tripped up by negatives or fractions while simplifying, remember: practice makes perfect! Double-check your work after each step and take your time with the arithmetic—it's super easy to miss a sign or make a little mistake that can lead you astray. And don't be afraid to rewrite the problem or backtrack if something doesn’t feel right. Algebra is like a puzzle, and every puzzle piece has its place!