How do you solve $3 x + y = 2$ $3x+y=2$ and $2 x + 3 y = 13$ $2x+3y=13$ using substitution?

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How do you solve $3 x + y = 2$ $3x+y=2$ and $2 x + 3 y = 13$ $2x+3y=13$ using substitution?

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Answer 1 · 2016-03-04T17:44:06

$Point of intersection (x, y) \to (- 1, 5)$ $"Point of intersection "(x,y)" "->" "(-1,5)$

Explanation:

Given:
$3 x + y = 2$ $3x+y=2$ ............................(1)
$2 x + 3 y = 13$ $2x+3y=13$ .......................(2)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Consider equation (1)

Subtract $3 x$ $3x$ form both sides

$3 x - 3 x + y = 2 - 3 x$ $3x-3x+y=2-3x$

$0+y=-3x+2" "..................(1_a)$

Using equation $1_a$ substitute for $y$ in equation (2)

$2x+3(-3x+2)=13................(2_a)$

$2x-9x+6=13$

$-7x+6=13$

Subtract 6 from both sides

$-7x+6-6=13-6$

$-7x+0=7$

Divide bot sides by -7

$(-7)/(-7)xx x=(+7)/(-7)$

$(+1)xx x=-1$

$color(red)(x=-1)$
'~~~~~~~~~~~~~~~~~~~~~~~~~
Substitute $x=-1$ into equation (1)

$3x+y=2" "->" "3(-1)+y=2$

$-3+y=2$

Add 3 to both sides

$-3+3+y=2+3$

$color(red)(y=5)$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
To prove this:
write equation (1) as: $y=-3x+2$
write equation (2) as: $y=-2/3 x+13/3$
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How do you solve $3 x + y = 2$ $3x+y=2$ and $2 x + 3 y = 13$ $2x+3y=13$ using substitution?

1 Answer

Explanation:

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How do you solve $3 x + y = 2$ $3x+y=2$ and $2 x + 3 y = 13$ $2x+3y=13$ using substitution?