Question

If f (x) = ax + b, where a and b are integers, f (–1) = – 5 and f (3) = 3, then a and b are equal to

Answer 1

Bonjour!!

f(x) = ax + b = -5

f(-1) = a(-1) + b = -5
-a + b = -5
-a = -5 -b
a = 5 + b


f(3) = a(3) + b = 3
3a + b = 3
3(5 + b) + b = 3
15 + 3b + b = 3
4b = 12
b = 3

a = 5 + b
a = 5 + 3
a = 8

Answer 2

Answer:

Value of a is 2 and value of b is -3.

Step-by-step explanation:

Given,

$f(x) = ax + b.....(1)$

where a and b are integers.

Here also given,$f( - 1) = - 5$

and $f(3) = 3$

By value putting we can solve this problem.

We are putting x=(-1)

$f( - 1) = ( - 1)a + b = b - a$

We know f(-1)= -5

So,

$b - a = - 5.....(2)$

Again we are putting x=3 and get,

$f(3) = 3a + b$

We know, f(3)=3

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

We are subtracting equation (3) from equation (2),

$(b - a) - (3a + b) = - 5 - 3 \\ b - a - 3a - b = - 8 \\ - 4a = - 8 \\ 4a = 8 \\ a = \frac{8}{4} \\ a = 2$

We are putting a=2 in equation (3),

$3 \times 2 + b = 3 \\ 6 + b = 3 \\ b = 3 - 6 \\ b = - 3$

This is a problem of Algebra.

Some important formulas of Algebra,

(a + b)² = a² + 2ab + b²

(a − b)² = a² − 2ab − b²

(a + b)³ = a³ + 3a²b + 3ab² + b³

(a - b)³ = a³ - 3a²b + 3ab² - b³

a³ + b³ = (a + b)³ − 3ab(a + b)

a³ - b³ = (a -b)³ + 3ab(a - b)

${a}^{2} - {b}^{2} = (a + b)(a - b)\\{a}^{2} + {b}^{2} = {(a + b)}^{2} - 2ab\\{a}^{2} + {b}^{2} = {(a - b)}^{2} + 2ab\\{a}^{3} - {b}^{3} = (a - b)( {a}^{2} + ab + {b}^{2} )\\{a}^{3} + {b}^{3} = (a + b)( {a}^{2} - ab + {b}^{2} )$