# If \frac{a}{b}=\frac{c}{d} Why does \frac{a+c}{b+d}=\frac{a}{b}=\frac{c}{d}?

If $$\displaystyle{\frac{{{a}}}{{{b}}}}={\frac{{{c}}}{{{d}}}}$$ Why does $$\displaystyle{\frac{{{a}+{c}}}{{{b}+{d}}}}={\frac{{{a}}}{{{b}}}}={\frac{{{c}}}{{{d}}}}?$$

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sonorous9n
Sketch: If you have $$\displaystyle{\frac{{{p}}}{{{q}}}}$$ and $$\displaystyle{\frac{{\lambda{p}}}{{\lambda{q}}}}$$, then
$$\displaystyle{\frac{{{p}+\lambda{p}}}{{{q}+\lambda{q}}}}={\frac{{{\left({1}+\lambda\right)}{p}}}{{{\left({1}+\lambda\right)}{q}}}}={\frac{{{p}}}{{{q}}}}$$
provided $$\displaystyle{1}+\lambda\ne{0}$$
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Donald Cheek
Consider $$\displaystyle{\frac{{{a}}}{{{b}}}}={\frac{{{k}{a}}}{{{k}{b}}}}$$, Then
$$\displaystyle{\frac{{{a}+{k}{a}}}{{{b}+{k}{b}}}}={\frac{{{\left({k}+{1}\right)}{a}}}{{{\left({k}+{1}\right)}{b}}}}={\frac{{{a}}}{{{b}}}}$$
which is exactly what you noticed, but with $$\displaystyle{a}={1},{b}={2},{k}={2}$$
Vasquez

We know ad=bc, so ab+bc=ab+ad. If you factor out this equation you get b(a+c)=a(b+d) and then you get $$\frac{a}{b}=\frac{a+c}{b+d}$$ . Similarly, you can prove the other