일차함수와 그래프 연습문제 (중학 수학)

401.OpenStax Intermediate Algebra · 3.2절 · 10번우리말 번역

기울기와 y절편을 이용하여 두 직선이 평행한지 판단하세요:

ⓐ $3x-2y=6$ 그리고 $y=\frac{3}{2}x+1$ ⓑ $y=2x-3$ 그리고 $-6x+3y=-9.$

영어 원문 보기

Use slopes and y-intercepts to determine if the lines are parallel:

ⓐ $3x-2y=6$ and $y=\frac{3}{2}x+1$ ⓑ $y=2x-3$ and $-6x+3y=-9.$

정답·풀이 보기

ⓐ


	$\begin{array}{cc}\quad 3x-2y & =6\end{array}$	그리고	$\quad \begin{array}{cc}y & =\frac{3}{2}x+1\end{array}$
첫 번째 방정식을 $y$ 에 대해 푼다.	$\quad \begin{array}{cc}-2y & =-3x+6 \\ \frac{-2y}{-2} & =\frac{-3x+6}{-2}\end{array}$
이제 이 방정식은 기울기–절편 꼴이다.	$\quad \begin{array}{cc}y & =\frac{3}{2}x-3\end{array}$
두 번째 직선의 방정식은 이미 기울기–절편 꼴이다.			$\quad \begin{array}{cc}y & =\frac{3}{2}x+1\end{array}$
두 직선의 기울기와 $y$ 절편을 각각 찾는다.	$\quad \begin{array}{cc}y & =\frac{3}{2}x-3 \\ y & =mx+b \\ m & =\frac{3}{2}\end{array}$		$\quad \begin{array}{cc}y & =\frac{3}{2}x+1 \\ y & =mx+b \\ y & =\frac{3}{2}\end{array}$
	$y\text{절편은}\,(0,-3)$		$\quad y\text{절편은}\,(0,1)$

두 직선은 기울기가 같고 y절편이 다르므로 평행하다.

두 직선이 평행한지 확인하려면 그래프를 그려 보는 것도 좋다.

ⓑ


	$\quad \begin{array}{cc}y & =2x-3\end{array}$	그리고	$\quad \begin{array}{cc}-6x+3y & =-9\end{array}$
첫 번째 방정식은 이미 기울기–절편 꼴이다.	$\quad \begin{array}{cc}y & =2x-3\end{array}$
두 번째 방정식을 $y$ 에 대해 푼다.			$\quad \begin{array}{cc}-6x+3y & =-9 \\ 3y & =6x-9 \\ \frac{3y}{3} & =\frac{6x-9}{3} \\ y & =2x-3\end{array}$
이제 두 번째 방정식은 기울기–절편 꼴이다.			$\quad \begin{array}{cc}y & =2x-3\end{array}$
두 직선의 기울기와 $y$ 절편을 각각 찾는다.	$\quad \begin{array}{cc}y & =2x-3 \\ y & =mx+b \\ m & =2\end{array}$		$\quad \begin{array}{cc}y & =2x-3 \\ y & =mx+b \\ m & =2\end{array}$
	$y\text{절편은}\,(0,-3)$		$\quad y\text{절편은}\,(0,-3)$

두 직선은 기울기가 같지만 y절편도 같다. 두 방정식은 같은 직선을 나타내며, 이러한 두 직선을 일치한다고 한다. 두 직선은 평행하지 않다. 같은 직선이다.

원본 교재 풀이(영어)

ⓐ


	$\begin{array}{cc}\quad 3x-2y & =6\end{array}$	and	$\quad \begin{array}{cc}y & =\frac{3}{2}x+1\end{array}$
Solve the first equation for $y.$	$\quad \begin{array}{cc}-2y & =-3x+6 \\ \frac{-2y}{-2} & =\frac{-3x+6}{-2}\end{array}$
The equation is now in slope–intercept form.	$\quad \begin{array}{cc}y & =\frac{3}{2}x-3\end{array}$
The equation of the second line is already in slope–intercept form.			$\quad \begin{array}{cc}y & =\frac{3}{2}x+1\end{array}$
Identify the slope and $y$ -intercept of both lines.	$\quad \begin{array}{cc}y & =\frac{3}{2}x-3 \\ y & =mx+b \\ m & =\frac{3}{2}\end{array}$		$\quad \begin{array}{cc}y & =\frac{3}{2}x+1 \\ y & =mx+b \\ y & =\frac{3}{2}\end{array}$
	$y\text{-intercept is}\,(0,-3)$		$\quad y\text{-intercept is}\,(0,1)$

The lines have the same slope and different y-intercepts and so they are parallel.

You may want to graph the lines to confirm whether they are parallel.

ⓑ


	$\quad \begin{array}{cc}y & =2x-3\end{array}$	and	$\quad \begin{array}{cc}-6x+3y & =-9\end{array}$
The first equation is already in slope–intercept form.	$\quad \begin{array}{cc}y & =2x-3\end{array}$
Solve the second equation for $y.$			$\quad \begin{array}{cc}-6x+3y & =-9 \\ 3y & =6x-9 \\ \frac{3y}{3} & =\frac{6x-9}{3} \\ y & =2x-3\end{array}$
The second equation is now in slope–intercept form.			$\quad \begin{array}{cc}y & =2x-3\end{array}$
Identify the slope and $y$ -intercept of both lines.	$\quad \begin{array}{cc}y & =2x-3 \\ y & =mx+b \\ m & =2\end{array}$		$\quad \begin{array}{cc}y & =2x-3 \\ y & =mx+b \\ m & =2\end{array}$
	$y\text{-intercept is}\,(0,-3)$		$\quad y\text{-intercept is}\,(0,-3)$

The lines have the same slope, but they also have the same y-intercepts. Their equations represent the same line and we say the lines are coincident. They are not parallel; they are the same line.

한 번 더 — 비슷한 문제 풀기OpenStax Prealgebra · 11.4절 · 15번


첫 번째 방정식은 기울기–절편 꼴이다.	$\quad y=-5x-4$
두 번째 방정식을 $y$ 에 대해 푼다.		$\quad \begin{array}{cc}x-5y & =5 \\ -5y & =-x+5 \\ \frac{-5y}{-5} & =\frac{-x+5}{-5} \\ y & =\frac{1}{5}x-1\end{array}$
각 직선의 기울기를 구한다.	$\begin{array}{cc}y & =-5x-4 \\ y & =mx+b \\ {m}_{1} & =-5\end{array}$	$\quad \begin{array}{cc}y & =\frac{1}{5}x-1 \\ y & =mx+b \\ {m}_{2} & =\frac{1}{5}\end{array}$


두 방정식을 $y$ 에 대해 푼다.	$\begin{array}{cc}7x+2y & =3 \\ 2y & =-7x+3 \\ \frac{2y}{2} & =\frac{-7x+3}{2} \\ y & =-\frac{7}{2}x+\frac{3}{2}\end{array}$	$\quad \begin{array}{cc}2x+7y & =5 \\ 7y & =-2x+5 \\ \frac{7y}{7} & =\frac{-2x+5}{7} \\ y & =-\frac{2}{7}x+\frac{5}{7}\end{array}$
각 직선의 기울기를 구한다.	$\quad \begin{array}{cc}y & =mx+b \\ {m}_{1} & =-\frac{7}{2}\end{array}$	$\quad \begin{array}{cc}y & =mx+b \\ {m}_{1} & =-\frac{2}{7}\end{array}$


The first equation is in slope–intercept form.	$\quad y=-5x-4$
Solve the second equation for $y.$		$\quad \begin{array}{cc}x-5y & =5 \\ -5y & =-x+5 \\ \frac{-5y}{-5} & =\frac{-x+5}{-5} \\ y & =\frac{1}{5}x-1\end{array}$
Identify the slope of each line.	$\begin{array}{cc}y & =-5x-4 \\ y & =mx+b \\ {m}_{1} & =-5\end{array}$	$\quad \begin{array}{cc}y & =\frac{1}{5}x-1 \\ y & =mx+b \\ {m}_{2} & =\frac{1}{5}\end{array}$


Solve the equations for $y.$	$\begin{array}{cc}7x+2y & =3 \\ 2y & =-7x+3 \\ \frac{2y}{2} & =\frac{-7x+3}{2} \\ y & =-\frac{7}{2}x+\frac{3}{2}\end{array}$	$\quad \begin{array}{cc}2x+7y & =5 \\ 7y & =-2x+5 \\ \frac{7y}{7} & =\frac{-2x+5}{7} \\ y & =-\frac{2}{7}x+\frac{5}{7}\end{array}$
Identify the slope of each line.	$\quad \begin{array}{cc}y & =mx+b \\ {m}_{1} & =-\frac{7}{2}\end{array}$	$\quad \begin{array}{cc}y & =mx+b \\ {m}_{1} & =-\frac{2}{7}\end{array}$