Question

Розв’яжіть систему рівнянь 1) х-4у=3 2)ху+2у=9

Answer 1

вот так вот))))))))))))))))))))))

Answer 2

х=3+4у

(3+4у)у+2у=9

3у+4у во 2 степени+2у=9

4у во 2 степени+ 5у=9

4у во 2 степени+ 5у-9=0

дискриминант=5в 2 степени-4*4*(-9)=25+144=169

упервое=-5+13/8=1

увторое=-5-13/8=-2.25

х=3+4=7

хвторое=3+4*(-2.25)=-6

 

 

Answer 3

Sure! I can help you solve the system of equations. Let's start by writing down the given equations:

Equation 1: x - 4y = 3

Equation 2: xy + 2y = 9

To solve this system of equations, we can use the method of substitution or elimination. I will show you how to solve it using the substitution method.

Step 1: Solve Equation 1 for x in terms of y. From Equation 1, we have: x = 3 + 4y

Step 2: Substitute the value of x in Equation 2. Substituting x = 3 + 4y in Equation 2, we get: (3 + 4y)y + 2y = 9

Simplifying this equation, we have: 3y + 4y^2 + 2y = 9 6y + 4y^2 = 9

Step 3: Rearrange the equation into a quadratic form. Rearranging the equation, we have: 4y^2 + 6y - 9 = 0

Step 4: Solve the quadratic equation. To solve this quadratic equation, we can use the quadratic formula: y = (-b ± √(b^2 - 4ac)) / (2a)

For our equation, a = 4, b = 6, and c = -9. Substituting these values into the quadratic formula, we have: y = (-6 ± √(6^2 - 4 * 4 * -9)) / (2 * 4)

Simplifying further, we get: y = (-6 ± √(36 + 144)) / 8 y = (-6 ± √180) / 8 y = (-6 ± 6√5) / 8

So we have two possible values for y: y1 = (-6 + 6√5) / 8 y2 = (-6 - 6√5) / 8

Step 5: Substitute the values of y back into Equation 1 to find the corresponding values of x. Substituting y = (-6 + 6√5) / 8 into Equation 1, we get: x - 4((-6 + 6√5) / 8) = 3 x - (-3/2 + 3√5/2) = 3 x + 3/2 - 3√5/2 = 3 x = 3 - 3/2 + 3√5/2 x = 3/2 + 3√5/2

Similarly, substituting y = (-6 - 6√5) / 8 into Equation 1, we get: x = 3/2 - 3√5/2

So the solution to the system of equations is: x = 3/2 + 3√5/2 y = (-6 + 6√5) / 8

and

x = 3/2 - 3√5/2 y = (-6 - 6√5) / 8

Please note that these are the approximate values, and you can simplify them further if needed.