Question

What are the solutions to the system of equations?

Answer 1

A

Explanation:

Given the 2 equations

y = x² - 4x + 8 → (1)

y = 2x + 3 → (2)

Since both express y in terms of x we can equate the left sides, that is

x² - 4x + 8 = 2x + 3 ( subtract 2x + 3 from both sides )

x² - 6x + 5 = 0 ← in standard form

(x - 1)(x - 5) = 0 ← in factored form

Equate each factor to zero and solve for x

x - 1 = 0 ⇒ x = 1

x - 5 = 0 ⇒ x = 5

Substitute these values into (2) for corresponding values of y

x = 1 : y = (2 × 1) + 3 = 2 + 3 = 5 ⇒ (1, 5)

x = 5 : y = (2 × 5) + 3 = 10 + 3 = 13 ⇒ (5, 13)

Solutions are (1, 5) and (5, 13)

Answer 2

A.
`(1,5)` and
`(5,13)`

Explanation:

We have been given a system of equations. We are asked to find the solution of our given system.

`\left \{ {{y=x^2-4x+8}\atop {y=2x+3}} \right.`

Equate both equations:

`x^2-4x+8=2x+3`

`x^2-4x-2x+8-3=2x-2x+3-3`

`x^2-6x+5=0`

Split the middle term:

`x^2-5x-x+5=0`

`x(x-5)-1(x-5)=0`

`(x-5)(x-1)=0`

Use zero product property:

`(x-5)=0\text{ (or) }(x-1)=0`

`x=5\text{ (or) }x=1`

Now, we will substitute both values of x, in our given equation to solve for y.

`y=2x+3`

`y=2(1)+3`

`y=2+3`

`y=5`

One solution:
`(1,5)`

`y=2x+3`

`y=2(5)+3`

`y=10+3`

`y=13`

2nd solution:
`(5,13)`

Therefore, option A is the correct choice.