Problem 24 - College Algebra

Question

Solve each system by the addition method. \(\left\\{\begin{array}{l}2 x-7 y-2 \\ 3 x+y--20\end{array}\right.\)

Step-by-Step Solution

Answer

The solutions to the system of equations are \((x,y) = (\sqrt{5}, 2\sqrt{2})\) and \((x,y) = (-\sqrt{5}, -2\sqrt{2})\)

Step 1: Identify the Equations

Identify the two equations from the given system:\[ 16x^2 - 4y^2 -72 = 0 \ x^2 - y^2 - 3 = 0 \]

Step 2: Multiply the Second Equation

Notice if we multiply the second equation by 4, then the coefficients in front of \(y^2\) will be the same in both equations. So let's multiply the second equation by 4, to get: \[ 4x^2 - 4y^2 -12 = 0 \] Now the system of equations will look like: \[\begin{array}{l} 16x^2 - 4y^2 -72 = 0 \ 4x^2 - 4y^2 -12 = 0 \end{array}\]

Step 3: Subtract the Equations

Now, subtract the second equation from the first one to eliminate \(y^2 \): \[ (16x^2 - 4y^2 -72) - (4x^2 - 4y^2 -12) = 0 \] This simplification yields: \[ 12x^2 - 60 = 0 \]

Step 4: Solving for x

Solve this equation for x: \[ x^2 = 5 \] Therefore, the solutions are \(x = \sqrt{5}\) and \(x = -\sqrt{5}\)

Step 5: Solving for y

Substitute \(x = \sqrt{5}\) and \(x = -\sqrt{5}\) in the second equation of the system to solve for \(y\). This gives, for \(y^2\), the values 8 and 8 respectively. So, \(y = \sqrt{8} = 2\sqrt{2}\) and \(y = -\sqrt{8} = -2\sqrt{2}\)