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Sagot :
The solution for the given system of equations is (3,5) and (-4,12).
Given two systems of equations i.e. x+y=8 and y=[tex]x^{2}[/tex]-4 and asked to solve them.
⇒x+y=8; let's assume this equation as A
⇒ y=[tex]x^{2}[/tex]-4; let's assume this equation as B
From equation A
⇒y=8-x
From equation B
⇒y=[tex]x^{2}[/tex]-4
Equating both the values of y
⇒8-x=[tex]x^{2}[/tex]-4
⇒[tex]x^{2}[/tex]-8-4+x=0
⇒[tex]x^{2}[/tex]-12+x=0
{solving quadratic equation}
⇒[tex]x^{2}[/tex]-12+4x-3x=0 { factorizing the quadratic equation}
⇒[tex]x^{2}[/tex]+4x-3x-12=0
⇒x(x+4)-3(x+4)=0
⇒(x-3)(x+4)=0
Therefore, the value's of x=3,-4
⇒if x=3
From equation A
⇒y=8-3=5
For x=3 ; y=5
⇒if x=-4
From equation A
⇒y=8-(-4)=12
For x=-4 ; y=12
Therefore,The solution for the given system of equations is (3,5) and (-4,12).
Learn more about equation here:
https://brainly.com/question/10413253
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