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How to prove that $ a + b (c + a') = a + b $? ($a'$ is not $a$)

I can see that this expression is valid by using a truth table, but I would like to prove it by using only algebra expressions. Any leads?

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Notice these rules (or identities) $$1+X = 1, \qquad X + YX = (1+Y)X = X, \qquad X+X' = 1$$

Then \begin{align}a + b(a'+c) &= (a + ba) + b(a'+c)\\ &= a+b(a+a'+c) \\&= a+b\end{align}

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    Can you move from a+b(c+a′) to a+b(a+a′+c) by adding a just like that? – 2017-02-05
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    Yes, because $a = a+ba$ as noted! – 2017-02-05
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    sorry I still dont see how you went from a+b(c+a′) to a+b(a+a′+c). Is there a step or two you could add to illustrate the transition? – 2017-02-05
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    OK, edited. You just substitute $a$ with $a+ba$, so that you get $a + ba + ba' + bc$. – 2017-02-05
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    Thanks for taking the time to edit your answer. It's clearer to me now. – 2017-02-05