Ensino Fundamental ⇒ Sistemas de equações do primeiro grau Tópico resolvido

[raquelcds]

[tex3]\begin{cases}
ax=by=cz \\
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{d}
\end{cases}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\begin{cases}
ax=by=cz \\
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{d}
\end{cases}[/tex3]

Resposta

---

[tex3]x=\frac{a+b+c}{ad}y=\frac{a+b+c}{bd}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]x=\frac{a+b+c}{ad}y=\frac{a+b+c}{bd}[/tex3]

[petras]

raquelcds,
x =
[tex3]\mathsf{x=\frac{by}{a}, y = \frac{ax}{b}, z = \frac{ax}{c}\\

\frac{1}{\frac{by}{a}}+\frac{1}{ \frac{ax}{b}}+\frac{1}{\frac{ax}{c}}=d
\rightarrow \\
\frac{a}{by}+\frac{b}{ax}+\frac{c}{ax}=d\rightarrow \\
by =ax\rightarrow \frac{a}{ax}+\frac{b}{ax}+\frac{c}{ax}=d\rightarrow \\
a+b+c=ax.d\therefore \boxed{\color{red}x = \frac{a+b+c}{ad}}\\
Da ~mesma~forma:\frac{a}{by}+\frac{b}{ax}+\frac{c}{ax}=d\rightarrow \\
\frac{a}{by}+\frac{b}{by}+\frac{c}{by}=d\rightarrow \\
a+b+c = d.by\therefore \boxed{\color{red}y = \frac{a+b+c}{bd}}\\


}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\mathsf{x=\frac{by}{a}, y = \frac{ax}{b}, z = \frac{ax}{c}\\

\frac{1}{\frac{by}{a}}+\frac{1}{ \frac{ax}{b}}+\frac{1}{\frac{ax}{c}}=d
\rightarrow \\
\frac{a}{by}+\frac{b}{ax}+\frac{c}{ax}=d\rightarrow \\
by =ax\rightarrow \frac{a}{ax}+\frac{b}{ax}+\frac{c}{ax}=d\rightarrow \\
a+b+c=ax.d\therefore \boxed{\color{red}x = \frac{a+b+c}{ad}}\\
Da ~mesma~forma:\frac{a}{by}+\frac{b}{ax}+\frac{c}{ax}=d\rightarrow \\
\frac{a}{by}+\frac{b}{by}+\frac{c}{by}=d\rightarrow \\
a+b+c = d.by\therefore \boxed{\color{red}y = \frac{a+b+c}{bd}}\\


}[/tex3]

[raquelcds]

Como 1/d se transformou em d ?

[petras]

raquelcds,

Creio que o enunciado esteja errado.

Da forma como está a solução seria [tex3]y = \frac{d(a+b+c)}{c}, x = \frac{d(a+b+c)}{a}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]y = \frac{d(a+b+c)}{c}, x = \frac{d(a+b+c)}{a}[/tex3]

Teste:
Se a=1, b=2, c=3, x=6, y=3, z = 2, d=1
[tex3]a.x =b.y=c.z\rightarrow1.6 = 2.3=3.2=6\\
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=d \rightarrow \frac{1}{6}+\frac{1}{3}+\frac{1}{2}=1\rightarrow \frac{6}{6}=1\\
x=\frac{a+b+c}{ad}=\frac{6}{1}=6, y = \frac{a+b+c}{bd}=\frac{6}{2}=3[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]a.x =b.y=c.z\rightarrow1.6 = 2.3=3.2=6\\
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=d \rightarrow \frac{1}{6}+\frac{1}{3}+\frac{1}{2}=1\rightarrow \frac{6}{6}=1\\
x=\frac{a+b+c}{ad}=\frac{6}{1}=6, y = \frac{a+b+c}{bd}=\frac{6}{2}=3[/tex3]

[tex3]a.x =b.y=c.z\rightarrow1.6 = 2.3=3.2=6\\
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{d} \rightarrow \frac{1}{6}+\frac{1}{3}+\frac{1}{2}=1\rightarrow \frac{6}{6}=1\\
x=\frac{d(a+b+c)}{c}=\frac{6}{3}=2\neq 6, y = \frac{d(a+b+c)}{a}=\frac{6}{1}=6\neq 3[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]a.x =b.y=c.z\rightarrow1.6 = 2.3=3.2=6\\
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{d} \rightarrow \frac{1}{6}+\frac{1}{3}+\frac{1}{2}=1\rightarrow \frac{6}{6}=1\\
x=\frac{d(a+b+c)}{c}=\frac{6}{3}=2\neq 6, y = \frac{d(a+b+c)}{a}=\frac{6}{1}=6\neq 3[/tex3]