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aceskillu
 one year ago
how do i solve 5x7y=9 4x+3y=33
aceskillu
 one year ago
how do i solve 5x7y=9 4x+3y=33

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aceskillu
 one year ago
Best ResponseYou've already chosen the best response.0this with it multiple choice

gypsy1274
 one year ago
Best ResponseYou've already chosen the best response.1I'm assuming that those are two separate equations that should look like this: \(5x  7y = 9\) and \(4x +3y = 33\) In that case, choose one equation (it doesn't matter which one) and solve for a variable (again, it doesn't matter which one). Then, substitute the resulting expression into the other equation in place of the variable that you have chosen. I'll provide an example: \(2x – 3y = –2\) and \(4x + y = 24\) I'm going to choose the second equation and solve for y. \(4x + y = 24 \) First, isolate y by subtracting both sides by 4x: \(y = 244x\) If y had a coefficient(other than 1) you would divide both sides by that number. It doesn't so we are done with this step. Now, Substitute \(244x\) for y in the other equation: \(2x3(244x) = 2\) And solve this equation using Order of Operations. Parenthesis: \(2x72+12x = 2\) Combine like terms by adding 72 to each side then combine the x terms: \(14x = 70\) Divide both sides by 14 to isolate the x: \(x = 5\) Now, substitute 5 into each equation and solve for y: (hint: you should get the same answer on each equation  this is how you know you are correct.) \(2x – 3y = –2\) and \(4x + y = 24\) \(2(5) – 3y = –2\) and \(4(5) + y = 24\) \(10 – 3y = –2\) and \(20 + y = 24\) \(3y = 12\) and \(y = 4\) \(y = 4\) If there is anything that is unclear, please ask.

texaschic101
 one year ago
Best ResponseYou've already chosen the best response.0good job gypsy :)

papsmurf
 one year ago
Best ResponseYou've already chosen the best response.0i dont see how you got 2x at the beginning of the problem @gypsy1274

gypsy1274
 one year ago
Best ResponseYou've already chosen the best response.1@papsmurf I don't like providing answers, so I created a different example to explain the steps. The original poster was already offline when I answered this question.
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