[tex] \large{ \begin{cases} y = 2x – 3 \\ y = – x + 3 \end{cases}}[/tex]
Since both are y-isolated equation, we can combine them together.
[tex] \large{2x – 3 = – x + 3}[/tex]
Isolate and solve for x-term.
[tex] \large{2x – 3 + 3 + x = – x + 3 + x + 3} \\ \large{2x + x = 6 \longrightarrow 3x = 6} \\ \large{ \frac{3x}{3} = \frac{6}{3} \longrightarrow \boxed{ \red{x = 2}}}[/tex]
Next, we find the value of y. Simply substitute x = 2 in one of these equations. The less coefficient values, the faster and better. I will substitute x = 2 in y = -x+3. You can substitute x = 2 in y = 2x-3 if you want but the result would be the same.
We are given the system of equations:
[tex] \large{ \begin{cases} y = 2x – 3 \\ y = – x + 3 \end{cases}}[/tex]
Since both are y-isolated equation, we can combine them together.
[tex] \large{2x – 3 = – x + 3}[/tex]
Isolate and solve for x-term.
[tex] \large{2x – 3 + 3 + x = – x + 3 + x + 3} \\ \large{2x + x = 6 \longrightarrow 3x = 6} \\ \large{ \frac{3x}{3} = \frac{6}{3} \longrightarrow \boxed{ \red{x = 2}}}[/tex]
Next, we find the value of y. Simply substitute x = 2 in one of these equations. The less coefficient values, the faster and better. I will substitute x = 2 in y = -x+3. You can substitute x = 2 in y = 2x-3 if you want but the result would be the same.
[tex] \large{y = – x + 3}[/tex]
Substitute x = 2 in the equation.
[tex] \large{y = – 2 + 3} \\ \large \boxed{ \blue{y = 1}}[/tex]
Therefore – when x = 2, y = 1. We can write in coordinate point or ordered pair as (2,1) from (x,y).
Answer