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Question abc=baia=ic+5abc=baia=ic+5\newlineA=b−cA=b-c\newlineB=14B=14\newlineC=b+aC=b+a\newlineSolve for ii

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Q. Question abc=baia=ic+5abc=baia=ic+5\newlineA=b−cA=b-c\newlineB=14B=14\newlineC=b+aC=b+a\newlineSolve for ii
  1. Simplify Equations: First, let's simplify the given equations. We have A=b−cA=b-c, B=14B=14, and C=b+aC=b+a. Since B=14B=14, we can substitute BB for bb in the other equations.
  2. Express aa and cc: Now we have A=14−cA=14-c and C=14+aC=14+a. We need to find expressions for aa and cc in terms of AA and CC to substitute into the equation abc=baia=i(c+5)abc=ba^{i}a=i(c+5).
  3. Substitute into abc=baia=ic+5abc=baia=ic+5: From A=14−cA=14-c, we can express cc as c=14−Ac=14-A. Similarly, from C=14+aC=14+a, we can express aa as a=C−14a=C-14.
  4. Simplify Equation: Substitute aa and cc into the equation abc=baia=ic+5abc=ba^ia=ic+5. We get (14)(C−14)(14−A)=(14−A)(14)(C−14)i=(14−A)(C−14)+5(14)(C-14)(14-A)=(14-A)(14)(C-14)i=(14-A)(C-14)+5.
  5. Solve for i: Simplify the equation. We get 196(C−14)−196(C−14)A=196(C−14)i−196A(C−14)i+5196(C-14)-196(C-14)A=196(C-14)i-196A(C-14)i+5.
  6. Correct Mistake: Now, let's solve for ii. We can divide both sides by 196(C−14)196(C-14) to isolate ii. We get 1−A=i−Ai+5196(C−14)1-A=i-Ai+\frac{5}{196(C-14)}.
  7. Correct Mistake: Now, let's solve for ii. We can divide both sides by 196(C−14)196(C-14) to isolate ii. We get 1−A=i−Ai+5196(C−14)1-A=i-Ai+\frac{5}{196(C-14)}.Rearrange the terms to solve for ii. We get i(1+A)=1−A+5196(C−14)i(1+A)=1-A+\frac{5}{196(C-14)}.
  8. Correct Mistake: Now, let's solve for ii. We can divide both sides by 196(C−14)196(C-14) to isolate ii. We get 1−A=i−Ai+5196(C−14)1-A=i-Ai+\frac{5}{196(C-14)}.Rearrange the terms to solve for ii. We get i(1+A)=1−A+5196(C−14)i(1+A)=1-A+\frac{5}{196(C-14)}.Divide both sides by (1+A)(1+A) to get i=1−A+5196(C−14)1+Ai=\frac{1-A+\frac{5}{196(C-14)}}{1+A}.
  9. Correct Mistake: Now, let's solve for ii. We can divide both sides by 196(C−14)196(C-14) to isolate ii. We get 1−A=i−Ai+5196(C−14)1-A=i-Ai+\frac{5}{196(C-14)}.Rearrange the terms to solve for ii. We get i(1+A)=1−A+5196(C−14)i(1+A)=1-A+\frac{5}{196(C-14)}.Divide both sides by (1+A)(1+A) to get i=1−A+5196(C−14)1+Ai=\frac{1-A+\frac{5}{196(C-14)}}{1+A}.Now we can plug in the values for AA and CC to find ii. But we made a mistake in the previous step; we should have simplified the equation before dividing by (1+A)(1+A). Let's go back and correct that.

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