How do you find the point of intersection for $x + y = 3$ $x+y=3$ and $2 x - y = - 3$ $2x-y= -3$ ?

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How do you find the point of intersection for $x + y = 3$ $x+y=3$ and $2 x - y = - 3$ $2x-y= -3$ ?

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Answer 1 · 2015-06-26T19:27:21

The given lines intersect at $(0, 3)$ $(0,3)$

Explanation:

[1] $XXXX$ $color(white)("XXXX")$ $x + y = 3$ $x+y=3$
[2] $XXXX$ $color(white)("XXXX")$ $2 x - y = - 3$ $2x-y= -3$

add [1] and [2]
[3] $XXXX$ $color(white)("XXXX")$ $3 x = 0$ $3x = 0$
[4] $XXXX$ $color(white)("XXXX")$ $x = 0$ $x=0$

substituting $0$ $0$ for $x$ $x$ in [1]
$XXXX$ $color(white)("XXXX")$ $0 + y = 3$ $0 + y = 3$
$XXXX$ $color(white)("XXXX")$ $y = 3$ $y=3$

Answer 2 · 2015-06-26T19:31:13

I found the point of intersection of coordinates:
$x = 0$ $x=0$
$y = 3$ $y=3$

Explanation:

You basically solve the System of the two equations trying to find values of $x$ $x$ and $y$ $y$ that satisfy both equations simultaneously.
From the first you can isolate $x$ $x$ as:
$x = 3 - y$ $x=3-y$
now you can substitute this $x$ $x$ into the second equation and find $y$ $y$ as:
$2 (3 - y) - y = - 3$ $2(3-y)-y=-3$
$6 - 2 y - y = - 3$ $6-2y-y=-3$
$- 3 y = - 9$ $-3y=-9$
$y = 3$ $y=3$
substitute back this value into the first equation to find $x$ $x$ :
$x = 3 - 3 = 0$ $x=3-3=0$

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How do you find the point of intersection for $x + y = 3$ $x+y=3$ and $2 x - y = - 3$ $2x-y= -3$ ?

2 Answers

Explanation:

Explanation:

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How do you find the point of intersection for $x + y = 3$ $x+y=3$ and $2 x - y = - 3$ $2x-y= -3$ ?