Question

semester 1: unit 2 exam review

1. what is the solution to the given system of equations?

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$$\begin{cases} 3x - 2y = 10 \\\\ x + y = 5 \\end{cases}$$

Explanation:

Step1: Isolate $x$ from second equation

From $x + y = 5$, we get $x = 5 - y$.

Step2: Substitute $x$ into first equation

Substitute $x=5-y$ into $3x - 2y = 10$:
$$3(5 - y) - 2y = 10$$
Expand and simplify:
$$15 - 3y - 2y = 10$$
$$15 - 5y = 10$$

Step3: Solve for $y$

Rearrange to solve for $y$:
$$-5y = 10 - 15$$
$$-5y = -5$$
$$y = \frac{-5}{-5} = 1$$

Step4: Substitute $y$ to find $x$

Substitute $y=1$ into $x = 5 - y$:
$$x = 5 - 1 = 4$$

Answer:

$x=4$, $y=1$ (or the ordered pair $(4, 1)$)