Answer:
[tex]\large \boxed{\sf \ \ 6 \ and \ 11 \ \ }[/tex]
Step-by-step explanation:
Hello,
Let's note a and b the two numbers.
a = b - 5
[tex]a^2+b^2=157[/tex]
We replace a in the second equation and we solve it
[tex](b-5)^2+b^2=157 \\ \\ \text{*** develop the expression ***} \\ \\b^2-10b+25+b^2=157 \\ \\ \text{*** subtract 157 from both sides ***} \\ \\2b^2-10b+25-157=2b^2-10b-132=0 \\ \\ \text{*** divide by 2 both sides ***} \\ \\b^2-5b-66=0[/tex]
It means that the sum of the two roots is 5 and the product is -66.
because
[tex](x-\alpha )(x-\beta )=x^2-(\alpha +\beta )x+\alpha \beta \\ \\ \text{ where } \alpha \text{ and } \beta \text{ are the roots }[/tex]
And we can notice that 66 = 6 * 11 and 11 - 6 = 5
So let's factorise it !
[tex]b^2-5b-66=0 \\ \\b^2+6b-11b-66=0 \\ \\b(b+6)-11(b+6)=0 \\ \\(b-11)(b+6) =0 \\ \\ b=11 \ or \ b=-6[/tex]
It means that the solutions are
(6,11) and (-6,-11)
I guess we are after positive numbers though.
Hope this helps.
Do not hesitate if you need further explanation.
Thank you
