Question

Can I get some help with this?

Answer 1

Solving the given system of equations we get solution set: {(4,0),(0,4)}

Option B is correct.

Step-by-step explanation:

We need to find the solution set of the following equations:

$x+y=4\\x^2+y^2=16$

Let:

$x+y=4\,\,\,eq(1)\\x^2+y^2=16\,\,\,eq(2)$

Finding value of x from eq(1) and putting in equation (2)

$x=4-y\\Putting\,\,in\,\,eq(2):\\(4-y)^2+y^2=16\\(16-8y+y^2)+y^2=16\\16-8y+y^2+y^2-16=0\\2y^2-8y=0\\Using\,\,quadratic\,\,equation\,\,to\,\,find\,\,value\,\,of\,\,y:\\y=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\a=2,\,\,b=-8\,\,,c=0\\Putting\,\,values:\\y=\frac{8\pm\sqrt{64}}{4}\\y=\frac{8\pm8}{4}\\y=\frac{8+8}{4}\,\,and\,\,y=y=\frac{8-8}{4}\\y=4\,\,and\,\,y=0$

Putting values of y=4 and y=0 in equation(1), to find values of x:

Putting y=4

$x+y=4\\x+4=4\\x=4-4\\x=0$

So, one solution is (0,4)

Putting y=0

$x+y=4\\x+0=4\\x=4$

So, other solution is: (4,0)

Hence Solving the given system of equations we get solution set: {(4,0),(0,4)}

Option B is correct.

Keywords: System of equations

Learn more about system of equations at:

• brainly.com/question/9045597
• brainly.com/question/3739260
• brainly.com/question/13168205

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