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rusak2 [61]
3 years ago
5

Solving multi step equations

Mathematics
1 answer:
hichkok12 [17]3 years ago
6 0

The no-brain way to do it is to

  1. subtract one side from both sides so you have <em>(something) = 0</em>.
  2. divide by the coefficient of the variable.
  3. add the opposite of the constant.

Of course, at some point, you need to simplify the equation so you have something like

... ax + b = 0 . . . . . . where <em>a</em> and <em>b</em> are some constants that may be positive or negative

17) Subtract the right side.

... 10(x +3) -(-9x -4) -(x -5 +3) = 0

... 10x + 30 +9x +4 -x +5 -3 = 0 . . . . . eliminate parentheses

... x(10+9-1) +(30+4+5-3) = 0 . . . . . . . collect terms

... 18x +36 = 0 . . . . . . . . . . . . . . . . . . . simplified

... x + 2 = 0 . . . . . . . . . . . . . . . . . . . . . .divide by 18, the coefficient of x

... x = -2 . . . . . . . . . . . . . . . . . . . . . . . . add the opposite of the constant

19) Add the opposite of the left side.

... 0 = -9(1 +7x) +12(x -12)

... 0 = -9 -63x +12x -144 . . . . . . eliminate parenthses

... 0 = -51x -153 . . . . . . . . . . . . . simplify

... 0 = x +3 . . . . . . . . . . . . . . . . . divide by -51, the coefficient of x

... -3 = x . . . . . . . . . . . . . . . . . . . add the opposite of the constant

_____

If you examine the variable's coefficients you can make a choice of side to subtract that results in a positive coefficient of the variable.

This method puts variable and constant together until the end. The approach usually taught is to separate the variable terms and constant terms. (The number of steps required is the same either way.)

The reason this is "no brain" is that it always works and requires no judgment as to what you add or subtract from where. Applying a little judgment as described above can make it so you're mostly working with positive numbers, but the method works whether the numbers are positive or negative.

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\begin{matrix}1\\1&1\\1&2&1\\1&3&3&1\\1&4&6&4&1\end{bmatrix}

The rows add up to 1,2,4,8,16, respectively. (Notice they're all powers of 2)

The sum of the numbers in row n is 2^{n-1}.

The last problem can be solved with the binomial theorem, but I'll assume you don't take that for granted. You can prove this claim by induction. When n=1,

(1+x)^1=1+x=\dbinom10+\dbinom11x

so the base case holds. Assume the claim holds for n=k, so that

(1+x)^k=\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k

Use this to show that it holds for n=k+1.

(1+x)^{k+1}=(1+x)(1+x)^k
(1+x)^{k+1}=(1+x)\left(\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k\right)
(1+x)^{k+1}=1+\left(\dbinom k0+\dbinom k1\right)x+\left(\dbinom k1+\dbinom k2\right)x^2+\cdots+\left(\dbinom k{k-2}+\dbinom k{k-1}\right)x^{k-1}+\left(\dbinom k{k-1}+\dbinom kk\right)x^k+x^{k+1}

Notice that

\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!}{\ell!(k-\ell)!}+\dfrac{k!}{(\ell+1)!(k-\ell-1)!}
\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)}{(\ell+1)!(k-\ell)!}+\dfrac{k!(k-\ell)}{(\ell+1)!(k-\ell)!}
\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)+k!(k-\ell)}{(\ell+1)!(k-\ell)!}
\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(k+1)}{(\ell+1)!(k-\ell)!}
\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{(k+1)!}{(\ell+1)!((k+1)-(\ell+1))!}
\dbinom k\ell+\dbinom k{\ell+1}=\dbinom{k+1}{\ell+1}

So you can write the expansion for n=k+1 as

(1+x)^{k+1}=1+\dbinom{k+1}1x+\dbinom{k+1}2x^2+\cdots+\dbinom{k+1}{k-1}x^{k-1}+\dbinom{k+1}kx^k+x^{k+1}

and since \dbinom{k+1}0=\dbinom{k+1}{k+1}=1, you have

(1+x)^{k+1}=\dbinom{k+1}0+\dbinom{k+1}1x+\cdots+\dbinom{k+1}kx^k+\dbinom{k+1}{k+1}x^{k+1}

and so the claim holds for n=k+1, thus proving the claim overall that

(1+x)^n=\dbinom n0+\dbinom n1x+\cdots+\dbinom n{n-1}x^{n-1}+\dbinom nnx^n

Setting x=1 gives

(1+1)^n=\dbinom n0+\dbinom n1+\cdots+\dbinom n{n-1}+\dbinom nn=2^n

which agrees with the result obtained for part (c).
4 0
3 years ago
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