How do you solve the system $y = 2 x + 5$ $y= 2x + 5$ and $y = 3 x - 8$ $y= 3x - 8$ by substitution?

Question

How do you solve the system $y = 2 x + 5$ $y= 2x + 5$ and $y = 3 x - 8$ $y= 3x - 8$ by substitution?

1 Answer

Impact of this question

2005 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-05-28T19:12:31

You substitute the $y$ $y$ -value of one equation as expressed in $x$ $x$ into the other.

In this case we get:
$2 x + 5 = 3 x - 8$ $2x+5=3x-8$ , subtract $2 x$ $2x$ from both sides
$\to 5 = x - 8$ $->5=x-8$ , add $8$ $8$ to both sides
$\to 13 = x \to x = 13$ $->13=x->x=13$

Science

Math

Humanities

... and beyond

How do you solve the system $y = 2 x + 5$ $y= 2x + 5$ and $y = 3 x - 8$ $y= 3x - 8$ by substitution?

1 Answer

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you solve the system y= 2x + 5y=2x+5y= 2x + 5 and y= 3x - 8y=3x−8y= 3x - 8 by substitution?

1 Answer

Related questions

Impact of this question

How do you solve the system $y = 2 x + 5$ $y= 2x + 5$ and $y = 3 x - 8$ $y= 3x - 8$ by substitution?