To solve the system of equations, we need to find the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] that satisfy both equations simultaneously. The given system of equations is:
[tex]\[
\begin{array}{l}
y = 2x + 1 \quad (1) \\
-4x + 2y = 2 \quad (2)
\end{array}
\][/tex]
First, we can use substitution to solve this system. We already have [tex]\( y \)[/tex] expressed in terms of [tex]\( x \)[/tex] from equation (1):
[tex]\[ y = 2x + 1 \][/tex]
Now, we substitute this expression for [tex]\( y \)[/tex] into equation (2):
[tex]\[ -4x + 2y = 2 \][/tex]
Substitute [tex]\( y = 2x + 1 \)[/tex] into the equation:
[tex]\[ -4x + 2(2x + 1) = 2 \][/tex]
Next, distribute the 2 through the parentheses:
[tex]\[ -4x + 4x + 2 = 2 \][/tex]
Combine like terms:
[tex]\[ 0x + 2 = 2 \][/tex]
This simplifies to:
[tex]\[ 2 = 2 \][/tex]
This statement is always true, which implies that there are an infinite number of solutions. This occurs when the two equations are essentially the same line, meaning they overlap completely.
To confirm this, let’s write both equations in a more comparable form:
- The first equation is already in slope-intercept form, [tex]\( y = 2x + 1 \)[/tex].
- Rewrite the second equation in the same form by isolating [tex]\( y \)[/tex]:
[tex]\[ -4x + 2y = 2 \][/tex]
Add [tex]\( 4x \)[/tex] to both sides:
[tex]\[ 2y = 4x + 2 \][/tex]
Divide by 2:
[tex]\[ y = 2x + 1 \][/tex]
Both equations reduce to [tex]\( y = 2x + 1 \)[/tex], confirming that they are the same line.
Since the two equations represent the same line, any point on this line will be a solution. Therefore, the system has infinite solutions.
Thus, the answer is:
[tex]\[ \text{d. Infinite solutions} \][/tex]
