Determine which relation is a function. question 13 options: a) {(3, 0), (– 2, – 2), (7, – 2), (– 2, 0)} b) c) y = 15x + 2 y = 1 5 x + 2 d)

x=3−2d,5,−2(1+d),5,−27−2d,5,−2(2+d),5,2(y−d),5

Step-by-step explanation:Solving for x. Want to solve for y or solve for d instead?

1 Simplify  0-20−2  to  -2−2.

3,-2,-27,-2-2,02y=1,5x+2d3,−2,−27,−2−2,02y=1,5x+2d

2 Simplify  -2-2−2−2  to  -4−4.

3,-2,-27,-4,02y=1,5x+2d3,−2,−27,−4,02y=1,5x+2d

3 Subtract 2d2d from both sides.

3-2d,-2-2d,-27-2d,-4-2d,02y-2d=1,5x3−2d,−2−2d,−27−2d,−4−2d,02y−2d=1,5x

4 Divide both sides by 1,51,5.

\frac{3-2d}{1},5,\frac{-2-2d}{1},5,\frac{-27-2d}{1},5,\frac{-4-2d}{1},5,\frac{02y-2d}{1},5=x

1

3−2d

,5,

1

−2−2d

,5,

1

−27−2d

,5,

1

−4−2d

,5,

1

02y−2d

,5=x

5 Factor out the common term 22.

\frac{3-2d}{1},5,\frac{-2(1+d)}{1},5,\frac{-27-2d}{1},5,\frac{-4-2d}{1},5,\frac{02y-2d}{1},5=x

1

3−2d

,5,

1

−2(1+d)

,5,

1

−27−2d

,5,

1

−4−2d

,5,

1

02y−2d

,5=x

6 Factor out the common term 22.

\frac{3-2d}{1},5,\frac{-2(1+d)}{1},5,\frac{-27-2d}{1},5,\frac{-2(2+d)}{1},5,\frac{02y-2d}{1},5=x

1

3−2d

,5,

1

−2(1+d)

,5,

1

−27−2d

,5,

1

−2(2+d)

,5,

1

02y−2d

,5=x

7 Factor out the common term 22.

\frac{3-2d}{1},5,\frac{-2(1+d)}{1},5,\frac{-27-2d}{1},5,\frac{-2(2+d)}{1},5,\frac{2(y-d)}{1},5=x

1

3−2d

,5,

1

−2(1+d)

,5,

1

−27−2d

,5,

1

−2(2+d)

,5,

1

2(y−d)

,5=x

8 Simplify  \frac{3-2d}{1}

1

3−2d

  to  (3-2d)(3−2d).

3-2d,5,\frac{-2(1+d)}{1},5,\frac{-27-2d}{1},5,\frac{-2(2+d)}{1},5,\frac{2(y-d)}{1},5=x3−2d,5,

1

−2(1+d)

,5,

1

−27−2d

,5,

1

−2(2+d)

,5,

1

2(y−d)

,5=x

9 Simplify  \frac{-2(1+d)}{1}

1

−2(1+d)

  to  (-2(1+d))(−2(1+d)).

3-2d,5,-2(1+d),5,\frac{-27-2d}{1},5,\frac{-2(2+d)}{1},5,\frac{2(y-d)}{1},5=x3−2d,5,−2(1+d),5,

1

−27−2d

,5,

1

−2(2+d)

,5,

1

2(y−d)

,5=x

10 Simplify  \frac{-27-2d}{1}

1

−27−2d

  to  (-27-2d)(−27−2d).

3-2d,5,-2(1+d),5,-27-2d,5,\frac{-2(2+d)}{1},5,\frac{2(y-d)}{1},5=x3−2d,5,−2(1+d),5,−27−2d,5,

1

−2(2+d)

,5,

1

2(y−d)

,5=x

11 Simplify  \frac{-2(2+d)}{1}

1

−2(2+d)

  to  (-2(2+d))(−2(2+d)).

3-2d,5,-2(1+d),5,-27-2d,5,-2(2+d),5,\frac{2(y-d)}{1},5=x3−2d,5,−2(1+d),5,−27−2d,5,−2(2+d),5,

1

2(y−d)

,5=x

12 Simplify  \frac{2(y-d)}{1}

1

2(y−d)

  to  (2(y-d))(2(y−d)).

3-2d,5,-2(1+d),5,-27-2d,5,-2(2+d),5,2(y-d),5=x3−2d,5,−2(1+d),5,−27−2d,5,−2(2+d),5,2(y−d),5=x

13 Switch sides.

x=3-2d,5,-2(1+d),5,-27-2d,5,-2(2+d),5,2(y-d),5x=3−2d,5,−2(1+d),5,−27−2d,5,−2(2+d),5,2(y−d),5

Done

is only a line? i think so

they can both be solutions depends on the equation

step-by-step explanation:

what is the equation? ?