Solve the equation using the addition method (fraction) 15/7 x + 5y = -5 3x + 7y = -7
\(\large\color{slate}{ \frac{\LARGE 15}{\LARGE 7} x + 5y = -5 }\) \(\large\color{slate}{ 3x + 7y = -7 }\) 1) Divide the first equation by 5. 2) multiply the first equation times -7. 3) add the equations. (this is going to eliminate the y for you
so what is 15/7 divided by 5?
i just get a long decimal
@SolomonZelman ????
15 / 5 =?
3
so 3/7?
yes, if you divide 15/7 by 5 you get 3/7
ok
so your new first equation, after dividing it by 5 will be?
so it would be 3/7 + y = -1 divide that by -7 and it's....
? + -7y = 7
3/7 divided by -7 is____?
@SolomonZelman ?
you get after dividing by 5, 3/7 x + y = -1 now mltiply times -7
3 + -7y = 7
now what?
3/7 * -7 = -3 1 * -7 = -7 -1 * -7 = 7 So, 3/7 x * -7 = -3x y * -7 = -7y -1 * -7 = 7 So your 1st equation would now be: \(\large\color{teal}{ -3x - 7y = 7 }\) \(\large\color{teal}{ 3x + 7y = -7 }\)
Ok, what do I do once I have the first equation?
if you have multiplied the first equation times 7, instead of multiplying times -7, then it would be: \(\large\color{teal}{ 3x + 7y = -7 }\) and your second equation as it was, is, \(\large\color{teal}{ 3x + 7y = -7 }\)
both equations are same, no?
so the answer is just 3x + 7y = -7? I need two numbers in a ( , ) thingy
when two equations are same, the have infinite solutions.
Ohhh ok. Thank you so much for the help! :)
sure
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