Question

which condition indicates that a system of equations might have one unique solution?
a. the lines intersect at one point.
b. the coefficients of $x$ and $y$ are different in each equation.
c. the equations are identical.
d. the lines are parallel.

what is a solution to the system $y = x + 2$ and $y = -x + 6$?
a. $(4,2)$
b. $(1,3)$
c. $(4,8)$
d. $(2,4)$

Explanation:

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First Question:

Step1: Recall solution types for linear systems

A system of linear equations has one unique solution when the graphs (lines) of the equations intersect exactly once. Parallel lines (no intersection) have no solution, identical equations (coinciding lines) have infinitely many solutions, and differing coefficients alone do not guarantee a unique solution (they could still be parallel).
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Second Question:

Step1: Set equations equal to each other

Since $y = x + 2$ and $y = -x + 6$, set $x + 2 = -x + 6$.

Step2: Solve for $x$

$$\begin{align*} x + x &= 6 - 2\\ 2x &= 4\\ x &= \frac{4}{2} = 2 \end{align*}$$

Step3: Substitute $x=2$ to find $y$

Use $y = x + 2$: $y = 2 + 2 = 4$.
The solution is $(2,4)$.

Answer:

1. a. The lines intersect at one point.
2. d. (2,4)