Na fração a/b, o numerador vale um terço do denominador. Somando-se 10 ao numerador, a fração torna-se equivalente a 1. Determine o valor de a+b.
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Olá.
Com as informações que foram dadas, podemos criar o seguinte sistema:
![\begin{cases} \mathsf{1^{a}~Eq.:}&\mathsf{a=\dfrac{b}{3}}\\\\ \mathsf{2^{a}~Eq.:}&\mathsf{(a+10)\div b=1} \end{cases} \begin{cases} \mathsf{1^{a}~Eq.:}&\mathsf{a=\dfrac{b}{3}}\\\\ \mathsf{2^{a}~Eq.:}&\mathsf{(a+10)\div b=1} \end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D+%5Cmathsf%7B1%5E%7Ba%7D%7EEq.%3A%7D%26amp%3B%5Cmathsf%7Ba%3D%5Cdfrac%7Bb%7D%7B3%7D%7D%5C%5C%5C%5C+%5Cmathsf%7B2%5E%7Ba%7D%7EEq.%3A%7D%26amp%3B%5Cmathsf%7B%28a%2B10%29%5Cdiv+b%3D1%7D+%5Cend%7Bcases%7D)
Substituindo o valor de "a" na segunda equação pelo valor de "a" da primeira equação, podemos encontrar o valor de "b" e consequentemente o valor de "a". Vamos aos cálculos.
![\mathsf{(a+10)\div b=1}\\\\\\ \mathsf{\left(\dfrac{b}{3}+10\right)\div b=1}\\\\\\ \mathsf{\left(\dfrac{b}{3}+10\cdot\dfrac{3}{3}\right)\div b=1}\\\\\\ \mathsf{\left(\dfrac{b}{3}+\dfrac{30}{3}\right)\div b=1}\\\\\\ \mathsf{\left(\dfrac{b+30}{3}\right)\cdot\dfrac{1}{b}=1}\\\\\\ \mathsf{\dfrac{b+30}{3b}=1}\\\\\\ \mathsf{b+30=3b}\\\\\\ \mathsf{30=3b-b}\\\\\\ \mathsf{30=2b}\\\\\\ \mathsf{\dfrac{30}{2}=b}\\\\\\ \boxed{\mathsf{15=b}} \mathsf{(a+10)\div b=1}\\\\\\ \mathsf{\left(\dfrac{b}{3}+10\right)\div b=1}\\\\\\ \mathsf{\left(\dfrac{b}{3}+10\cdot\dfrac{3}{3}\right)\div b=1}\\\\\\ \mathsf{\left(\dfrac{b}{3}+\dfrac{30}{3}\right)\div b=1}\\\\\\ \mathsf{\left(\dfrac{b+30}{3}\right)\cdot\dfrac{1}{b}=1}\\\\\\ \mathsf{\dfrac{b+30}{3b}=1}\\\\\\ \mathsf{b+30=3b}\\\\\\ \mathsf{30=3b-b}\\\\\\ \mathsf{30=2b}\\\\\\ \mathsf{\dfrac{30}{2}=b}\\\\\\ \boxed{\mathsf{15=b}}](https://tex.z-dn.net/?f=%5Cmathsf%7B%28a%2B10%29%5Cdiv+b%3D1%7D%5C%5C%5C%5C%5C%5C+%5Cmathsf%7B%5Cleft%28%5Cdfrac%7Bb%7D%7B3%7D%2B10%5Cright%29%5Cdiv+b%3D1%7D%5C%5C%5C%5C%5C%5C+%5Cmathsf%7B%5Cleft%28%5Cdfrac%7Bb%7D%7B3%7D%2B10%5Ccdot%5Cdfrac%7B3%7D%7B3%7D%5Cright%29%5Cdiv+b%3D1%7D%5C%5C%5C%5C%5C%5C+%5Cmathsf%7B%5Cleft%28%5Cdfrac%7Bb%7D%7B3%7D%2B%5Cdfrac%7B30%7D%7B3%7D%5Cright%29%5Cdiv+b%3D1%7D%5C%5C%5C%5C%5C%5C+%5Cmathsf%7B%5Cleft%28%5Cdfrac%7Bb%2B30%7D%7B3%7D%5Cright%29%5Ccdot%5Cdfrac%7B1%7D%7Bb%7D%3D1%7D%5C%5C%5C%5C%5C%5C+%5Cmathsf%7B%5Cdfrac%7Bb%2B30%7D%7B3b%7D%3D1%7D%5C%5C%5C%5C%5C%5C+%5Cmathsf%7Bb%2B30%3D3b%7D%5C%5C%5C%5C%5C%5C+%5Cmathsf%7B30%3D3b-b%7D%5C%5C%5C%5C%5C%5C+%5Cmathsf%7B30%3D2b%7D%5C%5C%5C%5C%5C%5C+%5Cmathsf%7B%5Cdfrac%7B30%7D%7B2%7D%3Db%7D%5C%5C%5C%5C%5C%5C+%5Cboxed%7B%5Cmathsf%7B15%3Db%7D%7D)
Substituindo o valor de "b" na primeira equação teremos o valor de a.
![\mathsf{a=\dfrac{b}{3}}\\\\\\ \mathsf{a=\dfrac{15}{3}}\\\\\\ \boxed{\mathsf{a=5}} \mathsf{a=\dfrac{b}{3}}\\\\\\ \mathsf{a=\dfrac{15}{3}}\\\\\\ \boxed{\mathsf{a=5}}](https://tex.z-dn.net/?f=%5Cmathsf%7Ba%3D%5Cdfrac%7Bb%7D%7B3%7D%7D%5C%5C%5C%5C%5C%5C+%5Cmathsf%7Ba%3D%5Cdfrac%7B15%7D%7B3%7D%7D%5C%5C%5C%5C%5C%5C+%5Cboxed%7B%5Cmathsf%7Ba%3D5%7D%7D)
Agora, calculando " a + b ", teremos:
![\mathsf{a+b=5+15=20} \mathsf{a+b=5+15=20}](https://tex.z-dn.net/?f=%5Cmathsf%7Ba%2Bb%3D5%2B15%3D20%7D)
Com as informações que foram dadas, podemos criar o seguinte sistema:
Substituindo o valor de "a" na segunda equação pelo valor de "a" da primeira equação, podemos encontrar o valor de "b" e consequentemente o valor de "a". Vamos aos cálculos.
Substituindo o valor de "b" na primeira equação teremos o valor de a.
Agora, calculando " a + b ", teremos:
Qualquer dúvida, deixe nos comentários.
Bons estudos.Perguntas interessantes