Full solution

Q. Question

abc=baia=ic+5

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A=b-c

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B=14

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C=b+a

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Solve for

i

Simplify Equations: First, let's simplify the given equations. We have $A=b-c$ , $B=14$ , and $C=b+a$ . Since $B=14$ , we can substitute $B$ for $b$ in the other equations.
Express $a$ and $c$ : Now we have $A=14-c$ and $C=14+a$ . We need to find expressions for $a$ and $c$ in terms of $A$ and $C$ to substitute into the equation $abc=ba^{i}a=i(c+5)$ .
Substitute into $abc=baia=ic+5$ : From $A=14-c$ , we can express $c$ as $c=14-A$ . Similarly, from $C=14+a$ , we can express $a$ as $a=C-14$ .
Simplify Equation: Substitute $a$ and $c$ into the equation $abc=ba^ia=ic+5$ . We get $(14)(C-14)(14-A)=(14-A)(14)(C-14)i=(14-A)(C-14)+5$ .
Solve for i: Simplify the equation. We get $196(C-14)-196(C-14)A=196(C-14)i-196A(C-14)i+5$ .
Correct Mistake: Now, let's solve for $i$ . We can divide both sides by $196(C-14)$ to isolate $i$ . We get $1-A=i-Ai+\frac{5}{196(C-14)}$ .
Correct Mistake: Now, let's solve for $i$ . We can divide both sides by $196(C-14)$ to isolate $i$ . We get $1-A=i-Ai+\frac{5}{196(C-14)}$ .Rearrange the terms to solve for $i$ . We get $i(1+A)=1-A+\frac{5}{196(C-14)}$ .
Correct Mistake: Now, let's solve for $i$ . We can divide both sides by $196(C-14)$ to isolate $i$ . We get $1-A=i-Ai+\frac{5}{196(C-14)}$ .Rearrange the terms to solve for $i$ . We get $i(1+A)=1-A+\frac{5}{196(C-14)}$ .Divide both sides by $(1+A)$ to get $i=\frac{1-A+\frac{5}{196(C-14)}}{1+A}$ .
Correct Mistake: Now, let's solve for $i$ . We can divide both sides by $196(C-14)$ to isolate $i$ . We get $1-A=i-Ai+\frac{5}{196(C-14)}$ .Rearrange the terms to solve for $i$ . We get $i(1+A)=1-A+\frac{5}{196(C-14)}$ .Divide both sides by $(1+A)$ to get $i=\frac{1-A+\frac{5}{196(C-14)}}{1+A}$ .Now we can plug in the values for $A$ and $C$ to find $i$ . But we made a mistake in the previous step; we should have simplified the equation before dividing by $(1+A)$ . Let's go back and correct that.