Answer: 256,1
Step-by-step explanation:
[tex]\sqrt{x} - \sqrt{y} =15\\\sqrt{y} = 49-3\sqrt{x} \\\\\sqrt{x} -(49-3\sqrt{x} ) =15 \\so, x=256\\substitute \{x}\ :\\\\\sqrt{y} = 49-3\sqrt{256} =y=1[/tex]
so, 256=x
and, y=1
Hope this Helped!
Answer:
1, 4 or -1, -4
Step-by-step explanation:
Let the two numbers be x and such that x > y.
According to the first condition:
[tex]x^2 -y^2= 15[/tex]
[tex]\implies x^2 =y^2+15[/tex]......(1)
According to the second condition:
[tex]3x^2 +y^2= 49[/tex]
[tex]\implies 3(y^2+15) +y^2= 49[/tex]
(From equation 1)
[tex]\implies 3y^2+45 +y^2= 49[/tex]
[tex]\implies 4y^2 =49-45[/tex]
[tex]\implies 4y^2 =4[/tex]
[tex]\implies y^2 =\frac{4}{4}[/tex]
[tex]\implies y^2 =1[/tex]
[tex]\implies y =\pm 1[/tex]
When y = 1
[tex]\implies x^2 =(1)^2+15=1+15=16[/tex]
[tex]\implies x =\pm 4[/tex]
When y = -1
[tex]\implies x^2 =(-1)^2+15=1+15=16[/tex]
[tex]\implies x =\pm 4[/tex]
Thus, the required numbers are either 1, 4 or -1, -4
