Which line has a y-intercept of +2 and x-intercept of 4 ? A. y=-(4)/(3)x+2 B. 6x-10 y=20 C. x+2y=4 D. 8x+6y=12

Find y-intercept: To find the y-intercept of a line, we set x to 0 and solve for y. Let's start with option A: y = -(4/3)x + 2 If x = 0, then y = -(4/3)(0) + 2 = 2. This line has a y-intercept of +2.Check x-intercept for option A: Now, let's check if the x-intercept is 4 for option A. To find the x-intercept, we set y to 0 and solve for x. 0 = -(4/3)x + 2 Multiply both sides by 3 to get rid of the fraction: 0 = -4x + 6 Add 4x to both sides: 4x = 6 Divide both sides by 4: x = 6/4 x = 1.5 This is not equal to 4, so option A does not have an x-intercept of 4.Check y-intercept for option B: Let's move on to option B: 6x - 10y = 20 To find the y-intercept, set x to 0: 6(0) - 10y = 20 -10y = 20 Divide both sides by -10: y = -2 This line does not have a y-intercept of +2, so option B is not correct.Find y-intercept for option C: Now, let's check option C: x + 2y = 4 To find the y-intercept, set x to 0: 0 + 2y = 4 2y = 4 Divide both sides by 2: y = 2 This line has a y-intercept of +2.Check x-intercept for option C: Next, let's check if the x-intercept is 4 for option C. To find the x-intercept, we set y to 0: x + 2(0) = 4 x = 4 This line has an x-intercept of 4. Option C has both the correct y-intercept and x-intercept, so it is the correct answer.Final Answer: We do not need to check option D since we have already found the correct answer in option C.

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Which line has a y-intercept of +2 and x-intercept of 4 ?
A.
y=-(4)/(3)x+2
B.
6x-10 y=20
C.
x+2y=4
D.
8x+6y=12

Which line has a $y$ -intercept of + $2$ and $x$ -intercept of $4$ ? $\newline$ A. $y=-\frac{4}{3} x+2$ $\newline$ B. $6 x-10 y=20$ $\newline$ C. $x+2 y=4$ $\newline$ D. $8 x+6 y=12$

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Algebra 1

Reflections of functions

Full solution

Q. Which line has a

y

-intercept of +

2

and

x

-intercept of

4

\newline

y=-\frac{4}{3} x+2

\newline

6 x-10 y=20

\newline

x+2 y=4

\newline

8 x+6 y=12

Find y-intercept: To find the y-intercept of a line, we set $x$ to $0$ and solve for $y$ . Let's start with option A: $y = -\frac{4}{3}x + 2$ If $x = 0$ , then $y = -\frac{4}{3}(0) + 2 = 2$ . This line has a y-intercept of $+2$ .
Check x-intercept for option A: Now, let's check if the x-intercept is $4$ for option A. $\newline$ To find the x-intercept, we set $y$ to $0$ and solve for $x$ . $\newline$ $0 = -\left(\frac{4}{3}\right)x + 2$ $\newline$ Multiply both sides by $3$ to get rid of the fraction: $\newline$ $0 = -4x + 6$ $\newline$ Add $4x$ to both sides: $\newline$ $4x = 6$ $\newline$ Divide both sides by $4$ : $\newline$ $y$ $0$ $\newline$ $y$ $1$ $\newline$ This is not equal to $4$ , so option A does not have an x-intercept of $4$ .
Check y-intercept for option B: Let's move on to option B: $\newline$ $6x - 10y = 20$ $\newline$ To find the y-intercept, set $x$ to $0$ : $\newline$ $6(0) - 10y = 20$ $\newline$ $-10y = 20$ $\newline$ Divide both sides by $-10$ : $\newline$ $y = -2$ $\newline$ This line does not have a y-intercept of $+2$ , so option B is not correct.
Find y-intercept for option C: Now, let's check option C: $\newline$ $x + 2y = 4$ $\newline$ To find the y-intercept, set $x$ to $0$ : $\newline$ $0 + 2y = 4$ $\newline$ $2y = 4$ $\newline$ Divide both sides by $2$ : $\newline$ $y = 2$ $\newline$ This line has a y-intercept of $+2$ .
Check x-intercept for option C: Next, let's check if the x-intercept is $4$ for option C. $\newline$ To find the x-intercept, we set $y$ to $0$ : $\newline$ $x + 2(0) = 4$ $\newline$ $x = 4$ $\newline$ This line has an x-intercept of $4$ . $\newline$ Option C has both the correct y-intercept and x-intercept, so it is the correct answer.
Final Answer: We do not need to check option $D$ since we have already found the correct answer in option $C$ .