Question

I need help in doing question 1B please.​

Answer 1

a. (5,6) or (6,5)

b.

Explanation:

First, solve the first equation for y in terms of x

y=11−x

Then replace y in the second equation with this expression in terms of x

x(11−x)=30

Expand

11x−x2=30

Rearrange into “standard” quadratic equation form(*)

x2−11x+30=0

b.

--> 2x + y = 1

x^2 + y^2 = 1 (This is the Unit Circle for Trig.)

x^2 + (1 - 2x)^2 = 1

x^2 + 1 - 4x + 4x^2 = 1

5x^2 - 4x = 0

x(5x - 4) = 0

x = 0 or x = 4/5

y = 1 or y = 1 - 2(4/5) = -3/5

Solutions are (0,1) and (0.8,-0.6).

And solve using quadratic equation

x=11±112−(4)(30)√2=11±1√2={102,122}

So we now now that x is either 5 or 6. Plug these back into either of the original equations to find the other.

y=11−5=6

y=11−6=5

So the two solutions to your pair of equations are (5,6) and (6,5).

Answer 2

(0, 1 ) and (
`(4)/(5)`, -
`(3)/(5)` )

Explanation:

given the equations

2x + y = 1 ( subtract 2x from both sides )

y = 1 - 2x → (1)

x² + y² = 1 → (2)

substitute y = 1 - 2x into (2)

x² +(1 - 2x)² = 1 ← expand squared factor using FOIL

x² + 1 - 4x + 4x² = 1 ( simplify left side )

5x² - 4x + 1 = 1 ( subtract 1 from both sides )

5x² - 4x = 0 ← factor out x from each term on the left side

x(5x - 4) = 0

equate each factor to zero and solve for x

x = 0

5x - 4 = 0 ⇒ 5x = 4 ⇒ x =
`(4)/(5)`

substitute each of these values into (1) for corresponding values of y

x = 0 : y = 1 - 2(0) = 1 - 0 = 1 ⇒ (0, 1 )

x =
`(4)/(5)` : y = 1 - 2(
`(4)/(5)` ) = 1 -
`(8)/(5)` =
`(5)/(5)` -
`(8)/(5)` = -
`(3)/(5)` ⇒ (
`(4)/(5)`, -
`(3)/(5)` )

solutions are (0, 1 ) and (
`(4)/(5)`, -
`(3)/(5)` )