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### Course: Algebra (all content) > Unit 3

Lesson 1: Two-variable linear equations intro- Two-variable linear equations intro
- Solutions to 2-variable equations
- Worked example: solutions to 2-variable equations
- Solutions to 2-variable equations
- Completing solutions to 2-variable equations
- Complete solutions to 2-variable equations
- Solutions to 2-variable equations: substitution (old)

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# Solutions to 2-variable equations: substitution (old)

An old video of Sal checking whether (3,-4) is a solution of 5x+2y=7 by substituting x=3 and y=-4. Created by Sal Khan and Monterey Institute for Technology and Education.

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## Video transcript

Is 3 comma negative 4 a
solution to the equation 5x plus 2y is equal to 7? So they're saying, does x
equal 3, y equal negative 4, satisfy this equation, or
this relationship right here? So one way to do it is just
to substitute x is equal to 3 and y equals
negative 4 into this and see if 5 times x plus 2
times y does indeed equal 7. So we have 5 times 3
plus 2 times negative 4. This is equal to 15. 15 plus negative 8, which
does indeed equal 7. So it does satisfy the equation. So it is on the line. It is a solution. x equals 3, y equals negative 4
is a solution to this equation. So we've essentially
answered our question. It is. Now, another way to
do it, and I'm not going to go into
the details here, is you could actually
graph the line. So maybe the line might
look something like this, I'm not going to
do it in detail. And you see, if you have
a very good drawing of it, you see whether the
point lies on the line. If the point, when you graph
the point, does lie on the line, it would be a solution. If the point somehow ends
up not being on the line then you'd know it
isn't a solution. But to do this, you would have
to have a very good drawing and so you could very
precisely determine whether it's on the line. If you do the
substitution method, if you just substitute the
values into the equation and see if it comes
out mathematically, this will always be exact. So this is all we really
had to do in this example. So it definitely is a
solution to the equation.