If possible, solve the system of equations. Use any method. I will fan and metal

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If possible, solve the system of equations. Use any method. I will fan and metal

Algebra
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4x+9y=10 (1) 3x+5y=4 (2)
Use the substitution method. do you know how to do that?
Yes but the only examples I have are x with no number

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I dont get how to do it with a number in front of x.
\[4x+9y=10\]\[3x+5y=4\] For this you could do substitution, as the other poster suggested, or you could do elimination. I'll walk you through both :-)
Okay thank you
Let's try substitution first. Pick one of the equations, and solve it for \(y\). Or \(x\), whatever you like :-)
3x+5y=4 y=-3/5x+4/5
Okay. Now, take the other equation, and wherever you see \(y\), replace it with \(-\frac{3}{5}x+\frac{4}{5}\) What do you get?
besides some messy algebra? :-)
x=-2
So (-2,2)
Yes!
Another way to do this is called elimination. With elimination, we add or subtract the equations together to get an equation with fewer variables. Sometimes we can do this right away, other times we need to multiply one or both of the equations by some factor to get the elimination to take place. \[4x+9y=10\]\[3x+5y=4\]To do elimination, we want the coefficients of the variables in one of the columns to be equal but opposite. Let's multiply the first equation by -3, and the second equation by 4. \[(-3)4x + (-3)9y = (-3)10\]\[(4)3x+(4)5y=(4)4\] or\[-12x-27y=-30\]\[\ \ \ 12x+20y=\ \ 16\] Agreed?
K. I know how to do graphing, elimination and substitution. I just didn't have any examples with it in the version thats all.. Thanks for your help.
Ok. Yeah, you just solve for the substitution variable, doesn't matter if the coefficient is something other than the nice, easy 1 :-)
Okay. Thank you
Always happy to hear someone say "yeah, I know how to do that already!" — usually it is "I don't know how to do that" :-)

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