Question

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Match the systems of equations with their solution sets.
y + 12 = x2 + x
x + y = 3

y - 15 = x2 + 4x
x - y = 1

y + 5 = x2 - 3x
2x + y = 1

y- 6 = x2 – 3x
x + 2y = 2

y-17 = x2 - 9x
-x + y = 1

y - 15 = -x2 + 4x
x + y = 1
Solution Set
Linear-Quadratic System of Equations
{(-2,3), (7,-6)
{(-5, 8). (3,0)
{(-2,5),(3,-5)
{(2, 3), (8,9)

Answer 1

{(-2,3), (7,-6) } -----> y - 15 = -
`x^(2)` + 4x

x + y = 1

{(-5, 8). (3,0) } ----> y + 12 =
`x^(2)` + x

x + y = 3

{(-2,5),(3,-5) } -----> y + 5 =
`x^(2)` - 3x

2x + y = 1

{(2, 3), (8,9)} -----> y-17 =
`x^(2)` - 9x

-x + y = 1

Explanation:

The simplest way to find the answer is by solving all the equations and finding the value of x and y for each of them.

Solving the equations ->

y + 12 =
`x^(2)` + x

x + y = 3

Substitute y = 3-x from second equation to first and solving the quadratic equation obtained i.e solving
`x^(2)` + 2x -15 = 0 , we get values of x = -5 , 3 and the corresponding values of y by substituting values of x in second equation , y = 8, 0 respectively. So, solution matched = {(-5, 8). (3,0) }.

Similarly solving other equations using exactly the same method as above we get the following solutions,

For,

y - 15 =
`x^(2)` + 4x

x - y = 1

we don't get any integer solution for this hence it has no match.

For,

y + 5 =
`x^(2)` - 3x

2x + y = 1

we get x = 3,-2 and corresponding y = -5,5

So, solution is {(-2,5),(3,-5) }

For,

y- 6 =
`x^(2)` – 3x

x + 2y = 2

we again don't get any integer solution for x and y so this has no match.

For,

y-17 =
`x^(2)` - 9x

-x + y = 1

we get x = 2,8 and corresponding y = 3,9

So, {(2, 3), (8,9)} is the solution match.

Lastly for,

y - 15 = -
`x^(2)` + 4x

x + y = 1

we get x = -2,7 and corresponding y = 3,6

So, {(-2,3), (7,-6)} is the solution match.