Question

1. {a^2+b^2=100 {a+b=14 2. {a+b=49 {a^2+b^2=1681 1 Смотреть ответ ssirina ssirina 1. a=14-b (14-b)^2+b^2=100 196-28b+b^2+b^2=100 2b^2-28b+96=0 b^2-14b+48=0 b1=6 a1=8 b2=8 a2=6 2. a=49-b (49-b)^2+b^2=1681 2401-98b+b^2+b^2=1681 2b^2-98b+720=0 b^2-49b+360=0 b1=9 a1=40 b2=40 a2=9

Answer 1

The solutions to the given equations are as follows:

1. For the equations:

   • $a^2 + b^2 = 100$
   • $a + b = 14$

   By substituting $a = 14 - b$ into the first equation, we get: $(14 - b)^2 + b^2 = 100$ Expanding and simplifying: $196 - 28b + b^2 + b^2 = 100$ Rearranging and combining like terms: $2b^2 - 28b + 96 = 0$ Factoring the quadratic equation: $(b - 6)(b - 8) = 0$

   So, we have two possible solutions:

   • $b_1 = 6$ and $a_1 = 8$
   • $b_2 = 8$ and $a_2 = 6$

2. For the equations:

   • $a + b = 49$
   • $a^2 + b^2 = 1681$

   By substituting $a = 49 - b$ into the second equation, we get: $(49 - b)^2 + b^2 = 1681$ Expanding and simplifying: $2401 - 98b + b^2 + b^2 = 1681$ Rearranging and combining like terms: $2b^2 - 98b + 720 = 0$ Factoring the quadratic equation: $(b - 9)(b - 40) = 0$

   So, we have two possible solutions:

   • $b_1 = 9$ and $a_1 = 40$
   • $b_2 = 40$ and $a_2 = 9$

Therefore, the solutions for the given equations are:

1. $a = 8$, $b = 6$ or $a = 6$, $b = 8$
2. $a = 40$, $b = 9$ or $a = 9$, $b = 40$