# which of the following lines has a y-intercept of 5? a. 7x+2y-10=0 b. 6x-3y+15=0 c. 6x+7y-35=0i think it's all of them

*print*Print*list*Cite

### 5 Answers

a. 7x + 2y - 10 = 0

Let us rewrite usnig the intercept form:

7x + 2y = 10

Divide by 10:

==> x/(10/7) + y/(10/2) = 1

==> x/(10/7) + y/5 = 1

**Then y-intercept = 5**

b. 6x-3y+15=0

6x - 3y = -15

Divide by 15:

==> x/(-15/6) + y/(15/3) = 1

==> x/(-15/6) + y/5 = 1

**Then y-intercept = 5**

** **

c. 6x+7y-35=0

==> 6x + 7y = 35

==> x/(35/6) + y/(35/7) = 1

==> x/(35/6) + y/5 = 1

**Then y-intercept = 5**

**Then the answer is all of the above has y-intercept 5.**

To check which of the given lines has a y intercept of 5, we'll have to put the given equations into the standard form:

y = mx + n, where m represents the slope of the line and n represents the y intercept.

We'll start with the first equation:

7x+2y-10=0

To put it into the standard form, we'll have to isolate y to the left side. For this reason, we'll subtract 7x and add 10 both sides:

2y = -7x + 10

We'll divide by 2 both sides:

y = -7x/2 + 10/2

**y = -7x/2 + 5**

The y intercept is n = 5

So, the line from the point a) has the y intercept of 5.

b) 6x-3y+15=0

To put it into the standard form, we'll have to isolate y to the left side. For this reason, we'll subtract 6x and 15 both sides:

-3y = -6x - 15

We'll divide by 3 both sides:

**y = 2x + 5**

We notice that the second line has the y intercept of 5, also.

c) 6x+7y-35=0

To put it into the standard form, we'll have to isolate y to the left side. For this reason, we'll subtract 6x and add 35 both sides:

7y = -6x + 35

We'll divide by 7 both sides:

y = -6x/7 + 35/7

**y = -6x/7 + 5**

**We notice that the third line has the y intercept of 5, too.**

**So, all 3 given lines have the y intercept of 5.**

We shall write the equations in the double inter cept form:

x/a+y/b = 1, where a is x intercept and b is y intercept.

i)7x+2y -10 = 0

Re write the equation as below:

7x+2y = 10.

Divide by 10:

(7/10)x+(2/10)y = 1.

Rewrite the equation as:

x/(10/7) + y/(10/2) = 1

Therefore x intercept = 10/7 and y intercept = 10/2 = 5.

(ii)

6x-3y+15 = 0.

Re write the equation as

6x-3y = -15.

Divide by -15:

(6/-15)x + (-3/-15)y = 1

Rewrite the equation as:

x/ (-15/6) + y/(-15/-3) = 1

So xintercept = -15/6 = -2.5 and y intercept = -15/-3 = 5.

(iii)

6x+7y -35 = 0

Rewrite as:

6x+7y = 35

Divide by 35:

(6/35)x + (7/35)y = 1

Rewite the equation as:

x/(35/6) + y/(35/7) = 1

So x intercept = 35/6 and y intercept = 35/7 = 5.

So the equations 6x+3y-15 = 0 and 6x+7y-35 = 0 intecept y axis at y = 5.

y-intercept is the line where the y-coordinate of line where it intercepts the y-axis.

Thus we calculate the y-intercept of any line by substituting 0 for value of x in the equation of line, and finding the corresponding value of y.

Using this method y-intercepts of the three given lines are calculated below.

a. 7x + 2y - 10 = 0

Substituting value of x = 0:

7*0 + 2y - 10 = 0

==> 27 = 10

y = 10/2 = 5

b. 6x - 3y + 15 = 0

Substituting value of x = 0:

7*0 - 3y + 15 = 0

==> -3y = -15

y = -15/-3 = 5

c. 6x + 7y - 35 = 0

Substituting value of x = 0:

6*0 + 7y - 35 = 0

==> 7y = 35

y = 35/7 = 5

Answer:

All the given lines have y-intercept of 5.

The equation of a line can be expressed as y=mx+b where m is the slope and b is the y-intercept.

For the given equations:

a) 7x+2y-10=0

=>2y=-7x+10

=>y=(-7/2)x+5

Therefore the y-intercept is 5.

b) 6x-3y+15=0

=>3y=6x+15

=>y=2x+5

Therefore the y-intercept is 5.

c) 6x+7y-35=0

=>7y=-6x+35

=>y=(-6/7)x+5

Therefore the y-intercept is 5.

**Therefore the y-intercept is 5 for all the three lines. **