Question

solve the system by elimination. first, clear the denominators.
\\(\frac{7x}{2} + \frac{7y}{3} = \frac{49}{2}\\)
\\(\frac{x}{4} + \frac{y}{3} = \frac{11}{4}\\)
the solution set is \\(\square\\).
(simplify your answer. type an ordered pair.)

Explanation:

Step1: Clear denominators for first equation

Multiply the first equation \(\frac{7x}{2}+\frac{7y}{3}=\frac{49}{2}\) by 6 (LCM of 2,3) to get:
\(6\times\frac{7x}{2}+6\times\frac{7y}{3}=6\times\frac{49}{2}\)
\(21x + 14y = 147\) (simplify each term)

Step2: Clear denominators for second equation

Multiply the second equation \(\frac{x}{4}+\frac{y}{3}=\frac{11}{4}\) by 12 (LCM of 4,3) to get:
\(12\times\frac{x}{4}+12\times\frac{y}{3}=12\times\frac{11}{4}\)
\(3x + 4y = 33\) (simplify each term)

Step3: Prepare for elimination (scale equations)

Let’s label the new equations:
Equation (1): \(21x + 14y = 147\)
Equation (2): \(3x + 4y = 33\)

Multiply Equation (2) by 7 to make coefficients of \(x\) equal:
\(7\times(3x + 4y)=7\times33\)
\(21x + 28y = 231\) (Equation 3)

Step4: Eliminate \(x\) by subtraction

Subtract Equation (1) from Equation (3):
\((21x + 28y)-(21x + 14y)=231 - 147\)
\(14y = 84\)

Step5: Solve for \(y\)

Divide both sides by 14:
\(y=\frac{84}{14}=6\)

Step6: Substitute \(y = 6\) into Equation (2)

Substitute \(y = 6\) into \(3x + 4y = 33\):
\(3x + 4(6)=33\)
\(3x + 24 = 33\)
Subtract 24: \(3x = 9\)
Divide by 3: \(x = 3\)

Answer:

\((3, 6)\)