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Explain, in complete sentences, how you would use the substitution method to solve the following system of equations.
 one year ago
 one year ago
Explain, in complete sentences, how you would use the substitution method to solve the following system of equations.
 one year ago
 one year ago

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itsjustme_lolBest ResponseYou've already chosen the best response.1
2x – y = 5 x + 2y = 15
 one year ago

AylinBest ResponseYou've already chosen the best response.1
Are you familiar with the substitution method?
 one year ago

itsjustme_lolBest ResponseYou've already chosen the best response.1
sorta, not very good in math
 one year ago

AWarnBest ResponseYou've already chosen the best response.0
Solve either of the equation for x and then substitute that value for x into the other equation.
 one year ago

AWarnBest ResponseYou've already chosen the best response.0
Take the second equation and solve for x: x=152y then substitute this into the first equation. You get 2*(152y)y=5 then solve for y.
 one year ago

AylinBest ResponseYou've already chosen the best response.1
Ok. I'm not going to give you the specifics for your problem, but I can tell you what to do generally. When you have a system of equations, in order to solve via the substitution method you first isolate one of the variables in one equation. So for example if you had this as a system:\[ax+by=c\]\[dx+ey=f\]we could start by isolating x in the first equation:\[x=\frac{ by }{ a }+\frac{ c }{ a }\]Then you just take that and plug it into x in the other equation:\[d \times (\frac{ by }{ a }+\frac{ c }{ a })+ey=f\]And then you just solve that equation for y. And then once you know what y, you just plug that back into \[x=\frac{ by }{ a }+\frac{ c }{ a }\]and that gives you x also. Does that make sense?
 one year ago

itsjustme_lolBest ResponseYou've already chosen the best response.1
thats still confusing..:( im sorry..
 one year ago

AylinBest ResponseYou've already chosen the best response.1
That's ok. How 'bout I show you an example that's similar to your problem?
 one year ago

ktnguyen1Best ResponseYou've already chosen the best response.0
Solve for y in the 1st equation and plug y value in the 2nd equation. Or, solve for x in the 2nd one and plug that x value in the 1st.
 one year ago

AylinBest ResponseYou've already chosen the best response.1
So say we have the following system:\[3xy=7\]\[x+4y=1\]So the first step is to isolate one of the variables in one of the equations. I'm going to choose to isolate y in the first equation, since that looks easy. So now I would write:\[3x3xy=73x\]\[1 \times (y)= 1 \times (73x)\]\[y=3x7\]In a sentence, I would say that I chose the first equation, subtracted 3x from both sides of the equal sign, and then multiplied both sides by 1. Then in the second equation instead of y, I would substitute in this new value we found for y:\[x+4(3x7)=1\]\[x+12x28=1\]\[13x28+28=1+28\]\[13x=29\]\[x=\frac{ 29 }{ 13 }\]Again in a sentence I would say that I substituted the value of y I found using the first equation into the second equation and then expanded, which resulted in x+12x28=1. Then I would say that I added 28 to both sides and then divided by 13 in order to find x. So now we know what x is, but what is y? Well, remember that equation we found for y in terms of x earlier? We just plug our x value into it and that will give us y, like so:\[y=3(\frac{ 29 }{ 13 })7\]\[y=\frac{ 87 }{ 13 }\frac{ 91 }{ 13 }\]\[y=\frac{ 4 }{ 13 }\] Does that clear it up for you?
 one year ago
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