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Sagot :
Leave the variable in each equation in one side of the equation to find if it has a solution:
a.
[tex]\begin{gathered} \frac{a+4}{5+a}=1 \\ \\ Multiply\text{ both sides by \lparen5+a\rparen:} \\ \frac{a+4}{5+a}(5+a)=1(5+a) \\ \\ a+4=5+a \\ \\ Subtract\text{ a in both sides of the equation:} \\ a-a+4=5+a-a \\ 4=5 \end{gathered}[/tex]As 4 is not equal to 5, the equation has no solution.
b.
[tex]\begin{gathered} \frac{1+b}{1-b}=1 \\ \\ Multiply\text{ both sides by \lparen1-b\rparen:} \\ \frac{1+b}{1-b}(1-b)=1(1-b) \\ \\ 1+b=1-b \\ \\ Add\text{ b in both sides of the equation:} \\ 1+b+b=1-b+b \\ 1+2b=1 \\ \\ Subtract\text{ 1 in both sides of the equation:} \\ 1-1+2b=1-1 \\ 2b=0 \\ \\ Divide\text{ both sides of the equation by 2:} \\ \frac{2b}{2}=\frac{0}{2} \\ \\ b=0 \\ \\ Prove\text{ if b=0 is the solution:} \\ \frac{1+0}{1-0}=1 \\ \\ \frac{1}{1}=1 \\ \\ 1=1 \end{gathered}[/tex]The solution for the equation is b=0
c.
[tex]\begin{gathered} \frac{c-5}{5-c}=1 \\ \\ Multiply\text{ both sides by \lparen5-c\rparen:} \\ \frac{c-5}{5-c}(5-c)=1(5-c) \\ \\ c-5=5-c \\ \\ Add\text{ c in both sides of the equation:} \\ c+c-5=5-c+c \\ 2c-5=5 \\ \\ Add\text{ 5 in both sides of the equation:} \\ 2c-5+5=5+5 \\ 2c=10 \\ \\ Divide\text{ both sides by 2:} \\ \frac{2c}{2}=\frac{10}{2} \\ \\ c=5 \\ \\ Prove\text{ if c=5 is a solution:} \\ \frac{5-5}{5-5}=1 \\ \\ \frac{0}{0}=1 \\ \\ \frac{0}{0}\text{ is undefined} \end{gathered}[/tex]As the possible solution (c=5) makes the expression has a undefined part (0/0) it is not a solution.
The equation has no solution
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