Answer:
The two number are:
[tex]y=1,\:x=\frac{1}{4}[/tex]
Step-by-step explanation:
Given that one number is 4 times another number, the first equation is
[tex]y = 4x[/tex]
Next, the sum of their reciprocals is 5, so
[tex]\frac{1}{x}+\frac{1}{y}=5[/tex]
Now, we have the system of equations
[tex]\begin{bmatrix}y=4x\\ \frac{1}{x}+\frac{1}{y}=5\end{bmatrix}[/tex]
Let us solve the system of equations
[tex]\begin{bmatrix}y=4x\\ \frac{1}{x}+\frac{1}{y}=5\end{bmatrix}[/tex]
substitute y = 4x in the equation [tex]\frac{1}{x}+\frac{1}{y}=5[/tex]
[tex]\frac{1}{x}+\frac{1}{4x}=5[/tex]
Least Common Multiple of x, 4x: 4x
Adjust fractions based on L.C.M
[tex]\frac{4}{4x}+\frac{1}{4x}=5[/tex]
[tex]\frac{4+1}{4x}=5[/tex]
[tex]\:\frac{5}{4x}=5[/tex]
Multiply both sides by 4x
[tex]\frac{5}{4x}\cdot \:4x=5\cdot \:4x[/tex]
Simplify
[tex]5=20x[/tex]
switch sides
[tex]20x=5[/tex]
Divide both sides by 20
[tex]\frac{20x}{20}=\frac{5}{20}[/tex]
Simplify
[tex]x=\frac{1}{4}[/tex]
For y = 4x, substitute [tex]x=\frac{1}{4}[/tex]
[tex]y = 4x[/tex]
[tex]y=4\cdot \frac{1}{4}[/tex]
[tex]y=1[/tex]
Therefore, the two number are:
[tex]y=1,\:x=\frac{1}{4}[/tex]
Verification:
[tex]\frac{1}{x}+\frac{1}{y}=5[/tex]
substituting [tex]y=1,\:x=\frac{1}{4}[/tex]
[tex]\:\frac{1}{\frac{1}{4}}+\frac{1}{1}=\:5\:\:[/tex]
[tex]4+1=5[/tex]
[tex]5 = 5[/tex]
L.H.S = R.H.S
and
[tex]y = 4x[/tex]
substituting [tex]y=1,\:x=\frac{1}{4}[/tex]
[tex]y\:=\:4\left(\frac{1}{4}\right)[/tex]
[tex]y=1[/tex]
