Solve the system: 3x + 4y = 12 and x − y = 2

Solve the system: 3x + 4y = 12 and x − y = 2

• A. x = 2, y = 0
• B. x = 4, y = 0
• C. x = 3, y = 1
• D. x = 20/7, y = 6/7

Answer: (D)

Solving a system of linear equations by substitution: from the second equation, x equals y plus 2. Substitute that into the first equation: 3(x) + 4y = 12 becomes 3(y + 2) + 4y = 12. Simplifying gives 3y + 6 + 4y = 12, so 7y + 6 = 12, hence y = 6/7. Then x = y + 2 = 6/7 + 2 = 20/7. The pair (x, y) = (20/7, 6/7) satisfies both equations, since 3x + 4y = 12 and x − y = 2 when you plug these values in. The other given pairs fail to satisfy at least one equation, so they’re not solutions.

Solving a system of linear equations by substitution: from the second equation, x equals y plus 2. Substitute that into the first equation: 3(x) + 4y = 12 becomes 3(y + 2) + 4y = 12. Simplifying gives 3y + 6 + 4y = 12, so 7y + 6 = 12, hence y = 6/7. Then x = y + 2 = 6/7 + 2 = 20/7. The pair (x, y) = (20/7, 6/7) satisfies both equations, since 3x + 4y = 12 and x − y = 2 when you plug these values in. The other given pairs fail to satisfy at least one equation, so they’re not solutions.