Answer:
(1/2, -3/4)
Explanation:
You want to solve by substitution the equations ...
Observation
We notice that the term 2y shows up in both equations. That means we can use one of them to write an expression for 2y that we can use in the other equation.
Expression to substitute
Solving the second equation for 2y, we have ...
2y = 5x -4 . . . . . add 5x to both sides
Substitution
Using this expression for 2y in the first equation gives ...
15x +(5x -4) = 6
20x = 10 . . . . . . . . . add 4, collect terms
x = 1/2 . . . . . . . . divide by 20, simplify
Now, we can use our expression above to find y:
2y = 5x -4 = 5(1/2) -4
2y = 5/2 -8/2 = -3/2 . . . . simplify
y = -3/4 . . . . . . . . . . . . divide by 2
The solution is (x, y) = (1/2, -3/4).
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Additional comment
The graph confirms this solution.
You can substitute anything that is convenient. For example, we could have rewritten the first equation to be ...
-3(-5x) +2y = 6
Then we could solve the second equation for -5x: -5x = -2y -4 and make the substitution ...
-3(-2y -4) +2y = 6
8y +12 = 6 . . . . . simplify
8y = -6 . . . . . . . subtract 12
y = -6/8 = -3/4 . . . . divide by 8
You're probably told that you need an expression for one of the variables. Solving the second equation for y gives ...
2y = 5x -4 . . . . . . add 5x to the second equation
y = (5x -4)/2 . . . . . . expression for y
This always works. As we have seen, in some cases it is not necessary to go this far. It usually pays to look at what you have to work with before you develop a strategy for solving it.