Answer:
Infinitely Many Solutions
Step-by-step explanation:
We are given the system of equations:
[tex]\displaystyle \large{\begin{cases} x = 2y-5 \\ -3x=-6y+15 \end{cases}}[/tex]
Since the first equation is already in x-isolated form, we can use the first equation to substitute in the second equation.
Substitute x = 2y-5 in the second equation.
[tex]\displaystyle \large{-3(2y-5)=-6y+15}[/tex]
Then we solve the equation for y-value; distribute -3 in.
[tex]\displaystyle \large{-6y + 15=-6y+15}[/tex]
We have to combine like terms. First, add both sides by 6y.
[tex]\displaystyle \large{-6y + 6y+ 15=-6y + 6y+15} \\ \displaystyle \large{ 15=15}[/tex]
Looks like both sides are equal. If both sides are equal and no variables exist, the equation is always true for all real numbers.
[tex] \displaystyle \large{y \in \R}[/tex]
Since y is true for all real numbers. That means x is also true for all real numbers. That's because for x = 2y-5, we can substitute y = any real numbers and still get the x-value. Therefore:
[tex] \displaystyle \large{x \in \R}[/tex]
Since it's the system of equations where we find the intercept(s) of graphs, the solutions for x and y are all real numbers, that means both graphs are same and intersect each others infinitely.
Let's take a look at both equations again this time.
[tex]\displaystyle \large{\begin{cases} x = 2y-5 \\ -3x=-6y+15 \end{cases}}[/tex]
For the second equation, we can divide the whole equation by -3.
[tex]\displaystyle \large{\begin{cases} x = 2y-5 \\ \frac{ - 3x}{ - 3} = \frac{ - 6y}{ - 3} + \frac{15}{ - 3} \end{cases}} \\ \displaystyle \large{\begin{cases} x = 2y-5 \\ x = 2y - 5 \end{cases}}[/tex]
Notice that two equations are same and therefore, both graphs are same and intersect each others infinitely.
Hence, the solution to this answer is Infinitely Many Solutions
