Respuesta :

zrh2sfo

Answer:

x = 3 , y = 1

Step-by-step explanation:

Solve the following system:

{y = x - 2 | (equation 1)

y = 7 - 2 x | (equation 2)

Express the system in standard form:

{-x + y = -2 | (equation 1)

2 x + y = 7 | (equation 2)

Swap equation 1 with equation 2:

{2 x + y = 7 | (equation 1)

-x + y = -2 | (equation 2)

Add 1/2 × (equation 1) to equation 2:

{2 x + y = 7 | (equation 1)

0 x+(3 y)/2 = 3/2 | (equation 2)

Multiply equation 2 by 2/3:

{2 x + y = 7 | (equation 1)

0 x+y = 1 | (equation 2)

Subtract equation 2 from equation 1:

{2 x+0 y = 6 | (equation 1)

0 x+y = 1 | (equation 2)

Divide equation 1 by 2:

{x+0 y = 3 | (equation 1)

0 x+y = 1 | (equation 2)

Collect results:

Answer: {x = 3 , y = 1

Ben

[tex]\huge\boxed{\boxed{\bold{(3, 1)}}}[/tex]

[tex]\hrulefill[/tex]

I'm assuming you need to find the solution to this system of equations (where the lines intersect).

We can use the substitution method to solve this system. Take the value of [tex]y[/tex] from the second equation and substitute it into the first:

[tex]-2x+7=x-2[/tex]

Add [tex]2x[/tex] to both sides of the new equation:

[tex]7=3x-2[/tex]

Now add [tex]2[/tex] to both sides of the equation:

[tex]9=3x[/tex]

Divide both sides by [tex]3[/tex]:

[tex]\boxed{3}=x[/tex]

Now let's solve for [tex]y[/tex] by substituting the known value of [tex]x[/tex] into the first equation:

[tex]y=3-2[/tex]

Simplify using subtraction:

[tex]y=\boxed{1}[/tex]

This means our solution is:

[tex]\large\boxed{(3, 1)}[/tex]