Here's the question you clicked on:
ValentinaT
Can someone check my work?
I have to solve systems of equations, and I chose the elimination method, but for some reason the last step isn't working. b) 7x – 3y = 20 5x + 3y = 16 Elimination Method 7x – 3y = 20 5x + 3y = 16 So with the elimination method you want to manipulate one equation so that when you add the two equations together one will cancel out. You have: 7x – 3y = 20 5x + 3y = 16 Eliminate one variable. Since the sum of the coefficients of y is 0, add the equations to eliminate y. 7x – 3y = 20 5x + 3y = 16 12x = 36 Now we can find: x = 3 In order to solve for y, take the value for x and substitute it back into either one of the original equations. 5(3) + 3y = 16 15 + 3y = 16
Could you show me where i went wrong?
I cannot understand how many ques you have posted up there . Can you please post one ques at a time.
I only posted one question..? I just asked if you could show me where in the problem did I do something wrong.
What are the equations and what is your answer for that.
I am sorry i am having a hard time understanding what you hav done above.@ValentinaT
Okay. The equation is: 7x – 3y = 20 5x + 3y = 16 I used the elimination method of solving systems to solve it. So with the elimination method you want to manipulate one equation so that when you add the two equations together one will cancel out. You have: 7x – 3y = 20 5x + 3y = 16 Eliminate one variable. Since the sum of the coefficients of y is 0, add the equations to eliminate y. 7x – 3y = 20 5x + 3y = 16 12x = 36 Now we can find: x = 3 In order to solve for y, take the value for x and substitute it back into either one of the original equations. 5(3) + 3y = 16 15 + 3y = 16
your answer should be x=3 y=1/3
15+3y=16 1)subract both sides by 15 2)divide both sides by 3
There can be a fraction and you can always check by putting the (x,y) value in your orignal equation and checking If, LHS=RHS then your answer is correct :)