# What are a and b if the points (1,3) and (3,7) are on the line f(x) if a*f(x+2) + b*f(2-x) = 6x + 25

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If the points belong to the graph of f(x), their coordinates verify the expression of f(x).

f(1) = 3 and f(3) =7

We'll put x = 1 in the expression of condition from enunciation:

a*f(1+2) + b*f(2-1) = 6 + 25

a*f(3) + b*f(1) = 31

We'll substitute f(1) and f(3):

7a + 3b = 31 (1)

We'll put x = -1 in the expression of condition from enunciation:

a*f(-1+2) + b*f(2+1) = -6 + 25

a*f(1) + b*f(3) = 19

3a + 7b = 19 (2)

We'll multiply (1) by -3:

-3(7a + 3b) = -93

-21a - 9b = -93 (3)

We'll multiply (2) by 7:

7(3a + 7b) = 133

21a + 49b = 133 (4)

We'll add (3) + (4):

-21a - 9b + 21a + 49b = -93 + 133

We'll eliminate and combine like terms:

40b = 40

**b = 1**

7a + 3 = 31

7a = 28

**a = 4**

Since the points (1,3) and (3, 7) are on the line f(x) = ax+b.

Therefore the line is : y-3 = (7-3)/((3-1){x-1). Or

y -3 = 2(x-1)

y = 2x-1 +3 = 2x+1

y = 2x+1

Therefore f(x) = 2x+1.

f(x+2) = 2(x+2)+1 = 2x+5.

f(2-x) = 2(2-x)+1 = 5-x.

af(x+2) +b(2-x) = 6x+25.

af(2x+5) +bf(2-x) = 6x+25.

a(2x+5) +b(5-2x)= 6x+25

2ax+5a+5b-2bx = 6x+25

2(a-b)x +5(a+b) = 6x+25.

Equating coefficients of like terms on both sides, we get:

2(a-b) = 6

a+b = 3....(1)

5(a +b) = 25

a+b = 5...(2).

Adding (1) and(2), we get: 2a = 3+5 = 8, Or a = 8/2 = 4.

(2) - (1) gives: 2b = 5-3 = 2. So b = 2/2 = 1.

Therefore a = 4 qnd b = 1.