Problem 22 - College Algebra

Question

Solve each system by the addition method. \(\left\\{\begin{array}{l}3 x+2 y-14 \\ 3 x-2 y-10\end{array}\right.\)

Step-by-Step Solution

Answer

The solutions to the system of equations are (1, 2), (1, -2), (-1,2), (-1, -2).

Step 1: Rearrange the equations

The first rearrangement can be to make both equations look more similar. Multiply the second equation by 2: \[4x^{2}-2y^{2}=-4\]. So, now the system to solve should look like: \[\left\{\begin{array}{l} 3x^{2}-2y^{2}=-5 \\ 4x^{2}-2y^{2}=-4 \end{array}\right..\]

Step 2: Subtract second equation from the first

Subtract the second equation from the first will remove the \( y^2 \) term. This process looks like:\[\begin{align*}(3x^{2}-4x^{2}) - (2y^{2}-2y^{2}) & = -5 - (-4) \\-x^{2} & = -1\end{align*}\]This simplifies to \(x^{2}=1\).

Step 3: Solve for x

Now, take the square roots on both sides of \(x^{2}=1\). The solutions would be \(x=1\) and \(x=-1\).

Step 4: Substitute the values of x into original equations

Substitute \(x=1\) into the first original equation to solve for \(y\). The equation will be \(3(1)^2-2y^2=-5\), and simplifies to \(2y^2=3+5\), so \(y^2=4\). This gives \(y=2\) and \(y=-2\). Repeat this for \(x=-1\). So all together \(x=1, y=2\) and \(x=1, y=-2\) and \(x=-1, y=2\) and \(x=-1, y=-2\) are all solutions.