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itsjustme_lol Group Title

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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  1. itsjustme_lol Group Title
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    2x – y = 5 x + 2y = 15

    • one year ago
  2. Aylin Group Title
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    Are you familiar with the substitution method?

    • one year ago
  3. itsjustme_lol Group Title
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    sorta, not very good in math

    • one year ago
  4. AWarn Group Title
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    Solve either of the equation for x and then substitute that value for x into the other equation.

    • one year ago
  5. AWarn Group Title
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    Take the second equation and solve for x: x=15-2y then substitute this into the first equation. You get 2*(15-2y)-y=5 then solve for y.

    • one year ago
  6. Aylin Group Title
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    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
  7. itsjustme_lol Group Title
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    thats still confusing..:( im sorry..

    • one year ago
  8. Aylin Group Title
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    That's ok. How 'bout I show you an example that's similar to your problem?

    • one year ago
  9. itsjustme_lol Group Title
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    that might help

    • one year ago
  10. ktnguyen1 Group Title
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    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
  11. Aylin Group Title
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    So say we have the following system:\[3x-y=7\]\[x+4y=1\]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:\[3x-3x-y=7-3x\]\[-1 \times (-y)= -1 \times (7-3x)\]\[y=3x-7\]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:\[x+4(3x-7)=1\]\[x+12x-28=1\]\[13x-28+28=1+28\]\[13x=29\]\[x=\frac{ 29 }{ 13 }\]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+12x-28=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
  12. itsjustme_lol Group Title
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    Yes, Thankyou!!

    • one year ago
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