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Sagot :
The given system of equations is
[tex]\begin{gathered} y=x^2+3\rightarrow(1) \\ y=x+5\rightarrow(2) \end{gathered}[/tex]Substitute y in equation (1) by equation (2)
[tex]x+5=x^2+3[/tex]Switch the 2 sides
[tex]x^2+3=x+5[/tex]Subtract x from both sides
[tex]\begin{gathered} x^2-x+3=x-x+5 \\ x^2-x+3=5 \end{gathered}[/tex]Subtract 5 from both sides
[tex]\begin{gathered} x^2-x+3-5=5-5 \\ x^2-x-2=0 \end{gathered}[/tex]Factor the left side into 2 factors
[tex]\begin{gathered} x^2=(x)(x) \\ -2=(-2)(1) \\ (x)(-2)+(x)(1)=-2x+x=-x\rightarrow(middle\text{ term\rparen} \\ x^2-x-2=(x-2)(x+1) \end{gathered}[/tex]The factors are (x - 2) and (x + 1)
[tex](x-2)(x+1)=0[/tex]Equate each factor by 0 to find the values of x
[tex]\begin{gathered} x-2=0 \\ x-2+2=0+2 \\ x=2 \end{gathered}[/tex][tex]\begin{gathered} x+1=0 \\ x+1-1=0-1 \\ x=-1 \end{gathered}[/tex]Substitute the values of x in equation (2) to find the values of y
[tex]\begin{gathered} y=2+5=7 \\ y=-1+5=4 \end{gathered}[/tex]The solutions of the system of equations are
(2, 7) and (-1, 4)
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