To solve the system of equations given by:
[tex]\[
\left\{
\begin{array}{l}
y = 4x - 3 \\
20x - 5y = 15
\end{array}
\right.
\][/tex]
we can follow these steps:
### Step 1: Substitute the expression for [tex]\( y \)[/tex] from the first equation into the second equation
From the first equation, we have:
[tex]\[ y = 4x - 3 \][/tex]
Substitute this expression for [tex]\( y \)[/tex] into the second equation:
[tex]\[ 20x - 5(4x - 3) = 15 \][/tex]
### Step 2: Simplify and solve for [tex]\( x \)[/tex]
Expand and simplify the equation:
[tex]\[ 20x - 5(4x) + 5(3) = 15 \][/tex]
[tex]\[ 20x - 20x + 15 = 15 \][/tex]
[tex]\[ 15 = 15 \][/tex]
Since the terms involving [tex]\( x \)[/tex] cancel out and we are left with [tex]\( 15 = 15 \)[/tex], which is always true, this indicates that the equations are dependent and represent the same line. Therefore, there are infinitely many solutions that lie on the line [tex]\( y = 4x - 3 \)[/tex].
### Conclusion:
The system of equations does not have a unique solution; rather, it has infinitely many solutions because the second equation is just a multiple of the first equation. Any point [tex]\((x, y)\)[/tex] that satisfies [tex]\(y = 4x - 3\)[/tex] will be a solution to the system. For example, if we choose [tex]\(x = 1\)[/tex]:
[tex]\[ y = 4(1) - 3 = 1 \][/tex]
So, [tex]\((1, 1)\)[/tex] is one solution. Similarly, if [tex]\(x = 2\)[/tex]:
[tex]\[ y = 4(2) - 3 = 5 \][/tex]
So, [tex]\((2, 5)\)[/tex] is another solution. This continues for all [tex]\( x \)[/tex] values that satisfy the relationship [tex]\( y = 4x - 3 \)[/tex].
