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Algebra Difficulty 1.6 Junior Find the answer Canada

The integers $a$ , $b$ and $c$ satisfy the equations $a+5=b$ and $5+b=c$ and $b+c=a$ . The value of $b$ is

Pick one

Solutions — 2

Solution 1

Since $a+5=b$ , then $a = b-5$ .

Since $a=b-5$ and $c=5+b$ and $b+c=a$ , then $\begin{align*} b + (5+b) & = b-5 \\ 2b + 5 & = b - 5 \\ b & = -10\end{align*}$ (If $b=-10$ , then $a=b-5=-15$ and $c = 5+b = -5$ and $b+c = (-10) + (-5) = (-15) = a$ , as
required.)

Solution 2

Since $a+5=b$ , then $a = b-5$ .

Substituting $a=b-5$ and $c=5+b$ into $b+c=a$ , we obtain $\begin{align*} b + (5+b) & = b-5 \\ 2b + 5 & = b - 5 \\ b & = -10\end{align*}$ (If $b=-10$ , then $a=b-5=-15$ and $c = 5+b = -5$ and $b+c = (-10) + (-5) = (-15) = a$ , as
required.)

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