For questions 5-7, find the solution to the system of equations by using either graphing or substitution. 6. y=2x-1 and y=x+3 A. (4, 7) B. (7, 4) C. (-7, -4) D. Infinite Solutions
so you have $$ 2x-1=y\\ \color{blue}{x+3=y}\\ --------- $$
I believe the answer is A.... Is that correct?
well, what would you need in the "blue" or bottom equation, to have it substract from the top and get "x" by itself?
Idk... I'm not very knowledgeable about algebra.
well, I assume you were given this because it has already being covered by the textbook, and you're supposed to know it
otherwise, you shouldn't be doing it
Mhm. And I have been trying to figure out the solution my textbooks solution is just a little complicated for me to understand..
lemme retype the equations some
$$ 2x-1=y\\ \color{blue}{x\ +\ 3=y}\\ ---------\\ \color{blue}{x\ +\ ?=\ ?}\\ $$
you'd want the 2nd equation, the bottom one, to "turn into" something that when added to the top one, gives only one variable, and cancels out the other
so you may end up with something like 3x+5=0, or 0+8=y, or so
for this instance, so you can see it, try $$ 2x-1=y\\ \color{blue}{x\ +\ 3=y} \ \ \times (-1)\\ ---------\\ \ \ \ 2x-1=y\\ \color{blue}{-x-3=-y}\\ ---------\\ $$ ? - ? = ?
add them both "vertically", what do you get?
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