Find the value of y, AB, BC and AC.?

Question

If B is between A and C, AB=5y+4, BC=7y-3 and AC=20y-7. Find the value of y, AB, BC and AC.

Answer 1

AB + BC = AC
as you said; AB=5y+4, BC=7y-3 ,AC=20y-7

5y+4 + 7y-3 = 20y-7
5y+4 + 7y-3 -20y +7=0
(5y+7y -20y) + ( 4 - 3 +7) = 0
(-8y) + ( 8 ) =0
-8y +8 =0
-8y=-8
y= -8/-8 and y=+1

if y= +1 ;AB = 5y+4 = 5 * 1 +4 = 5+4 =9
if y= +1 ;BC = 7y-3 = 7 * 1-3 = 7-3 = 4
if y= +1 ;AC = 20y-7 = 20 * 1 -7 = 20 - 7=13

AB = 9
BC = 4
AC = 13

Answer 2

Assuming that points A, B, and C are in alphabetical order, AC would be your longest line, 20y - 7.

AB and BC comprise AC, therefore AB + BC = AC

(5y + 4) + (7y - 3) = 20y - 7

5y + 7y + 4 - 3 = 20y - 7

12y + 1 = 20y - 7

Subtract 12y from both sides, so

1 = 20y - 12y - 7; 1 = 8y - 7

Add 7 to both sides, so

8 = 8y

Divide both sides by 8, so y = 1

Plug this back in, and AB = 9 = (5 X 1) + 4

BC = 4 = (7 X 1) - 3

AC = 13 = (20 X 1) - 7

9 + 4 = 13, so y must be 1

Answer 3

If B is between A and C, then by the definition of betweenness,
AB + BC = AC

Since
AB = 5y + 4
BC = 7y - 3
AC = 20y - 7,

(5y + 4) + (7y - 3) = (20y - 7)
12y + 1 = 20y - 7
20y - 12y = 1 + 7
8y = 8
y = 1
AB = 5y + 4 = 5(1) + 4 = 5 + 4 = 9
BC = 7y - 3 = 7(1) - 3 = 7 - 3 = 4
AC = 20y - 7 = 20(1) - 7 = 20 - 7 = 13

Therefore,
y = 1
AB = 9
BC = 4
AC = 13

^_^

Answer 4

i assume AC is straight line and B is an intermediate point, so
AB+B=AC
=> 5y+4 +7y -3=20y -7
=> 12y+1 = 20y-7
=> 8 = 8y
=> y=1.


now AB=5x1 +4 =9
BC= 7x1 - 3 = 4
AC= 20x1 -7=13

9+4=13

allthe best

Answer 5

AB + BC = AC

(5y + 4) + (7y - 3) = 20y - 7
5y + 4 + 7y - 3 = 20y - 7
12y + 1 = 20y - 7
-8y = -8
y = 1

AB = 5(1) + 4 = 5 + 4 = 9
BC = 7(1) - 3 = 7 - 3 = 4
AC = 20(1) - 7 = 20 - 7 = 13

ANS :
AB = 9
BC = 4
AC = 13

Answer 6

Given a + b + c + d = 8 sq. each and each part (a + b + c + d)² = 8² a² + b² + c² + d² + 2ab + 2ac + 2ad + 2bc + 2bd + 2cd = sixty 4 a² + b² + c² + d² + 2(ab + ac + ad + bc + bd + cd) = sixty 4 a² + b² + c² + d² + 2(12) = sixty 4 a² + b² + c² + d² + 24 = sixty 4 a² + b² + c² + d² = 40 ........................ {{a million}} And given ab + ac + ad + bc + bd + cd = 12 ab + ac + bc + d(a + b + c) = 12 ........................... {{2}} From {{a million}}. The equation represents the sum of squares of all actual numbers (so a² > 0, b² > 0, c² > 0, d² > 0), and our purpose the following is to maximise the linked fee of d. So assume a = b = c = 0 (to maximise d). notwithstanding from {{2}}, if a = b = c, LHS = 0 ? 12. subsequently our first assumption is faux. Now assume that a = b = 0, and c ? 0, (to lessen the form of words in {{2}} after simplified) The simplified equation of {{a million}} and {{2}}, given our 2d assumption are c² + d² = 40 .................. {{3}} cd = 12 ...................... {{4}} From {{4}} c = 12/d replace c = 12/d into {{3}} (12/d)² + d² = 40 100 and forty four/d² + d² = 40 d? - 40d² + 100 and forty four = 0 d² = 36, 4 (by using quadratic formula) ==> the optimum fee of d is 6 (with a = 0, b = 0, c = 12/6 = 2, values of a, b, c are interexchangeable) ok, my answer is faulty. = (

Answer 7

AB + BC = AC

5y+4 + 7y-3 = 20y-7
12y + 1 = 20y - 7
12y - 12y + 1 = 20y - 12y - 7 (subtract 12y from both sides)
1+7 = 8y - 7 + 7 (add 7 to both sides)
8y = 8
y = 1

AB = 9
BC = 4
AC = AB + BC = 9 + 4 = 13

Answer 8

give me a hint