Question

Решите систему а)х+3у=5х^-ху+у^=73б) 2х+у=71/х + 1/у = 5/6

Answer 1

Х-37:'-_8-373738393+__)(

Answer 2

It appears that there are some typos and missing information in the systems of equations you provided. Specifically, there are missing exponents and equations with equal signs instead of equal signs with numbers. I'll assume you meant the following systems:

a) $\begin{cases} x + 3y = 5 \\ x^2 - xy + y^2 = 73 \end{cases}$

b) $\begin{cases} 2x + y = 71 \\ \frac{1}{x} + \frac{1}{y} = \frac{5}{6} \end{cases}$

Let's solve these systems one by one:

a) $\begin{cases} x + 3y = 5 \\ x^2 - xy + y^2 = 73 \end{cases}$

To solve this system, we can use substitution or elimination. Let's use substitution:

From the first equation, we have $x = 5 - 3y$.

Now, we'll substitute this value of $x$ into the second equation:

$(5 - 3y)^2 - (5 - 3y)y + y^2 = 73$

Solving this equation will give us the value(s) of $y$, and then we can find the corresponding values of $x$ using the first equation.

b) $\begin{cases} 2x + y = 71 \\ \frac{1}{x} + \frac{1}{y} = \frac{5}{6} \end{cases}$

To solve this system, we'll use elimination. Let's manipulate the second equation to eliminate the fractions:

Multiply both sides by $6xy$ to get rid of the denominators:

$6y + 6x = \frac{5xy}{6}$

Rearrange the equation:

$6x + 6y = \frac{5xy}{6}$

Now, let's solve this system using the first equation:

$2x + y = 71$

$6x + 6y = \frac{5xy}{6}$

Now we can proceed to solve these systems. If you can clarify the equations with proper exponents and equal signs, I'd be happy to help you solve them.