Higher Math Chapter 5.4 solve this 1: (2x+3)(y-1)=14, (x-3)(y-2)=-1

Higher Math Chapter 5.4 solve this 1: (2x+3)(y-1)=14, (x-3)(y-2)=-1

To solve this system of equations, we can use the method of substitution.

First, we can simplify the first equation by expanding the left-hand side:

2xy - 2x + 3y - 3 = 14

Next, we can isolate one of the variables in terms of the other. Let's isolate y from the first equation:

2xy + 3y = 2x + 17 y(2x + 3) = 2x + 17 y = (2x + 17)/(2x + 3)

Now we can substitute this expression for y into the second equation:

(x - 3)((2x + 17)/(2x + 3) - 2) = -1

Simplifying this equation, we get:

(x - 3)(2x - 1) = -2(2x + 3)

Expanding the left-hand side and simplifying, we get:

2x^2 - 7x + 9 = 0

We can solve this quadratic equation using the quadratic formula:

x = (7 ± √(7^2 - 4(2)(9))) / (2(2))

x = (7 ± √37) / 4

Now we can use these values of x to find the corresponding values of y. Substituting x = (7 + √37) / 4 into the equation we found for y earlier, we get:

y = (2((7 + √37) / 4) + 17) / (2((7 + √37) / 4) + 3) y = (7 + √37) / 2

Similarly, substituting x = (7 - √37) / 4 into the equation we found for y, we get:

y = (7 - √37) / 2

Therefore, the solutions to the system of equations are:

x = (7 + √37) / 4, y = (7 + √37) / 2

and

x = (7 - √37) / 4, y = (7 - √37) / 2

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