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Sagot :
[tex]\begin{gathered} y=\frac{1}{3x}+4\ldots\ldots\ldots\ldots\ldots equation\text{ 1} \\ y=2x-1\ldots\ldots\ldots\ldots\ldots\text{equation 2} \\ \end{gathered}[/tex]
Step 1: We equation 1 to equation 2
[tex]\begin{gathered} \frac{1}{3x}+4=2x-1 \\ mulltiply\text{ through by 3x, we have;} \\ 1+12x=6x^2-3x \\ 6x^2-3x-12x-1=0 \\ 6x^2-15x-1=0 \\ \end{gathered}[/tex]Step 2: We apply the quadratic formula to solve the equation above to get two pairs of solution for x;
[tex]\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \text{Where a= 6, b=-15 , c= -1} \\ x=\frac{-(-15)\pm\sqrt[]{(-15)^2-4(6)(-1)}}{2(6)} \\ x=-0.06\text{ or x = }2.56 \end{gathered}[/tex]So, we have gotten solutions for x, let's proceed for y;
Step 3: For each value of x, we will solve for a corresponding value for y;
[tex]\begin{gathered} \text{When x=-0.06;} \\ y=2x-1 \\ y=2(-0.06)-1 \\ y=-1.12 \\ \text{When x=2.56;} \\ y=2(2.56)-1 \\ y=4.12 \end{gathered}[/tex]Hence, the ordered pairs of the solution are (-0.06, -1.12) and (2.56, 4.12)
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