Question

1. declan is solving the linear system of equations shown. he wants to use substitution.

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$$\begin{cases} 5x + 3y = 22 \\\\ 4x - 2y = 16 \\end{cases}$$

which variable is the easiest to solve for?
\bigcirc the $x$ in the second equation
\bigcirc the $y$ in the first equation
\bigcirc the $y$ in the second equation
\bigcirc the $x$ in the first equation

Explanation:

Step1: Analyze each option

For a variable to be easy to solve for, the coefficient should be a factor of the other terms (preferably with a smaller absolute value or leading to simpler fractions). Let's check each equation:

Equation 1: \(5x + 3y = 22\)

• Solving for \(x\): \(5x=22 - 3y\Rightarrow x=\frac{22 - 3y}{5}\) (involves dividing by 5, which may lead to fractions)
• Solving for \(y\): \(3y=22 - 5x\Rightarrow y=\frac{22 - 5x}{3}\) (also involves dividing by 3, fractions likely)

Equation 2: \(4x - 2y = 16\)

• Solving for \(x\): \(4x=16 + 2y\Rightarrow x=\frac{16 + 2y}{4}=\frac{8 + y}{2}\) (dividing by 4, simplifies to a linear expression with a denominator of 2)
• Solving for \(y\): \(- 2y=16 - 4x\Rightarrow y=\frac{4x - 16}{2}=2x - 8\) (dividing by - 2, which simplifies nicely to a linear expression without fractions, just integer coefficients)

Now, compare the simplicity of solving for each variable:

• Solving for \(y\) in the second equation gives \(y = 2x-8\), which is a simple linear expression with integer coefficients.
• Solving for \(x\) in the second equation gives \(\frac{8 + y}{2}\) (still has a denominator, but not as simple as solving for \(y\) here)
• Solving for \(x\) or \(y\) in the first equation leads to denominators of 5 or 3, which are more likely to result in fractional coefficients when substituted.

Step2: Conclusion

The easiest variable to solve for is the \(y\) in the second equation because when we solve \(4x - 2y=16\) for \(y\), we get \(y = 2x - 8\) which is a simple linear expression with integer coefficients, making substitution straightforward.

Answer:

the \(y\) in the second equation (corresponding to the option: the \(y\) in the second equation)