Question

-4x y = 6
-5x - y = 21
Soolve by substitution

Answer 1

From
`\(-4xy = 6\)` ,
`\(y = (6)/(-4x)\)`. Substituting into \(-5x - y = 21\) creates a quadratic equation. Solutions:
`\(x \approx 3.638\)` with
`\(y \approx -0.412\)` and
`\(x \approx 0.048\)` with
`\(y \approx -31.25\)`.

Given equations:

1.
`\(-4xy = 6\)`

2.
`\(-5x - y = 21\)`

From equation (1), isolate \(y\) in terms of \(x\):

`\[ y = (6)/(-4x) \]`

Now substitute this expression for \(y\) into equation (2):

`\[ -5x - (6)/(-4x) = 21 \]`

To eliminate the denominator, multiply both sides by \(-4x\):

`\[ -20x^2 + 6 = -84x \]`

Rearrange terms to form a quadratic equation:

`\[ 20x^2 - 84x + 6 = 0 \]`

Apply the quadratic formula:

`\[ x = (-b \pm √(b^2 - 4ac))/(2a) \]`

For the equation
`\(20x^2 - 84x + 6 = 0\)`:

`\[ a = 20, \, b = -84, \, c = 6 \]`

Substitute these values into the quadratic formula:

`\[ x = (84 \pm √((-84)^2 - 4(20)(6)))/(2(20)) \]`

`\[ x = (84 \pm √(7056 - 480))/(40) \]`

`\[ x = (84 \pm √(6576))/(40) \]`

`\[ x = (84 \pm 81.095)/(40) \]`

Two possible solutions for \(x\):

`\[ x_1 = (84 + 81.095)/(40) \approx 3.638 \]`

`\[ x_2 = (84 - 81.095)/(40) \approx 0.048 \]`

Now substitute these \(x\) values into
`\(y = (6)/(-4x)\)` to determine the corresponding values of \(y\):

For
`\(x_1\)`:

`\[ y_1 = (6)/(-4(3.638)) \approx -0.412 \]`

For
`\(x_2\)`:

`\[ y_2 = (6)/(-4(0.048)) \approx -31.25 \]`

Hence, the solutions are approximately
`\(x \approx 3.638\)` with
`\(y \approx -0.412\)`and \(x
`\approx 0.048\)` with
`\(y \approx -31.25\)`.

complete the question

"Given the system of equations:

1. \(-4xy = 6\)

2. \(-5x - y = 21\)

Solve the system of equations by substitution. Show the step-by-step process of isolating one variable in terms of the other, substituting the expression, and determining the values of \(x\) and \(y\). Finally, present the solutions for \(x\) and \(y\)."