Question

Решить систему уравнений способом подстановки

Answer 1

X^2+2y=5
y=1+x

x^2+2*(1+x)-5=0
y=1+x

x^2+2+2x-5=0
x^2+2x-3=0
D1=(2*2)^2-1*(-3)=1+3=4
x1=-1-2,                         x2=-1+2
x1=-3,                            x2=1

x1=-3,                                                 x2=1
y1=1-3,                                               y2=1+1

x1=-3,                                                  x2=1
y1=-2                                                   y2=2

Answer 2

Solving a System of Equations by Substitution

To solve a system of equations using the method of substitution, we follow these steps:

1. Solve one equation for one variable in terms of the other variable. 2. Substitute the expression obtained in step 1 into the other equation. 3. Solve the resulting equation for the remaining variable. 4. Substitute the value obtained in step 3 back into one of the original equations to find the value of the other variable. 5. Check the solution by substituting the values of the variables into both original equations.

Let's solve the system of equations using the method of substitution.

The system of equations is as follows:

1) \(x + 2y = 8\) 2) \(3x + y = 1\)

We will solve equation 1 for \(x\) in terms of \(y\):

\(x = 8 - 2y\) Now, we substitute the expression for \(x\) obtained in step 1 into equation 2:

\(3(8 - 2y) + y = 1\)

Simplifying the equation, we get:

\(24 - 6y + y = 1\)

Combining like terms, we have:

\(-5y + 24 = 1\)

Next, we solve the resulting equation for \(y\):

\(-5y = 1 - 24\)

\(-5y = -23\)

Dividing both sides of the equation by -5, we find:

\(y = \frac{-23}{-5} = \frac{23}{5}\)

Now, we substitute the value of \(y\) back into equation 1 to find the value of \(x\):

\(x + 2\left(\frac{23}{5}\right) = 8\)

Multiplying through by 5 to eliminate the fraction, we get:

\(5x + 2(23) = 40\)

Simplifying the equation, we have:

\(5x + 46 = 40\)

Subtracting 46 from both sides of the equation, we find:

\(5x = 40 - 46\)

\(5x = -6\)

Dividing both sides of the equation by 5, we get:

\(x = \frac{-6}{5} = -\frac{6}{5}\)

Therefore, the solution to the system of equations is \(x = -\frac{6}{5}\) and \(y = \frac{23}{5}\).

To check the solution, we substitute the values of \(x\) and \(y\) into both original equations:

Equation 1: \(x + 2y = 8\)

\(-\frac{6}{5} + 2\left(\frac{23}{5}\right) = 8\)

Simplifying the equation, we have:

\(-\frac{6}{5} + \frac{46}{5} = 8\)

\(\frac{40}{5} = 8\)

\(8 = 8\)

The equation is true, so the solution satisfies the first equation.

Equation 2: \(3x + y = 1\)

\(3\left(-\frac{6}{5}\right) + \frac{23}{5} = 1\)

Simplifying the equation, we have:

\(-\frac{18}{5} + \frac{23}{5} = 1\)

\(\frac{5}{5} = 1\)

\(1 = 1\)

The equation is true, so the solution satisfies the second equation.

Therefore, the solution \(x = -\frac{6}{5}\) and \(y = \frac{23}{5}\) is correct.

I hope this helps! Let me know if you have any further questions.