What are the values of \(x\) and \(y\) in the system of equations \(2x + 3y = 6\) and \(x - y = 1\)?

What are the values of \(x\) and \(y\) in the system of equations \(2x + 3y = 6\) and \(x - y = 1\)?

• A. \(x = 1, y = 1\)
• B. \(x = 2, y = 0\)
• C. \(x = 3, y = 0\)
• D. \(x = 0, y = 2\)

Answer: (C)

To find the values of \(x\) and \(y\) in the system of equations given by \(2x + 3y = 6\) and \(x - y = 1\), we can use substitution or elimination. First, let's solve the second equation for \(x\): \[ x = y + 1 \] Now, we can substitute \(x\) in the first equation: \[ 2(y + 1) + 3y = 6. \] Distributing gives: \[ 2y + 2 + 3y = 6. \] Combining like terms results in: \[ 5y + 2 = 6. \] Subtracting 2 from both sides yields: \[ 5y = 4, \] and then dividing by 5 gives: \[ y = \frac{4}{5}. \] Now, substitute \(y = \frac{4}{5}\) back into the expression we found for \(x\): \[ x = \frac{4}{5} + 1 = \frac{4}{5} + \frac{5}{5} = \frac{9}{5}. \

To find the values of (x) and (y) in the system of equations given by (2x + 3y = 6) and (x - y = 1), we can use substitution or elimination.

First, let's solve the second equation for (x):

[

x = y + 1

]

Now, we can substitute (x) in the first equation:

[

2(y + 1) + 3y = 6.

]

Distributing gives:

[

2y + 2 + 3y = 6.

]

Combining like terms results in:

[

5y + 2 = 6.

]

Subtracting 2 from both sides yields:

[

5y = 4,

]

and then dividing by 5 gives:

[

y = \frac{4}{5}.

]

Now, substitute (y = \frac{4}{5}) back into the expression we found for (x):

[

x = \frac{4}{5} + 1 = \frac{4}{5} + \frac{5}{5} = \frac{9}{5}.

\