Question

What are the orded pair of the soultion for this system of equation

Answer 1

`(2,-3)\text{ ;(3,-3)}`

Step-by-step explanation

Step 1

`\begin{gathered} f(x)=x^2-5x+3 \\ f(x)=-3 \end{gathered}`

there is a value for y, that is the same for both equations, so

`\begin{gathered} y_1=y_2 \\ x^2-5x+3=-3\Rightarrow equation\text{ (3)} \end{gathered}`

solve equation (3)

`\begin{gathered} x^2-5x+3=-3 \\ \text{add 3 in both sides} \\ x^2-5x+3+3=-3+3 \\ x^2-5x+6=0 \\ x^2-5x+6=0\Rightarrow ax^2+bx+c=0 \end{gathered}`

hence, we can use the quadratic formula

`\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \text{replace} \\ x=\frac{+5\pm\sqrt[]{(-5)^2-4\cdot1\cdot6}}{2\cdot1} \\ x=\frac{+5\pm\sqrt[]{25-24}}{2} \\ x=\frac{+5\pm\sqrt[]{1}}{2} \\ x=(+5\pm1)/(2) \\ x_1=(5+1)/(2)=(6)/(2)=3 \\ x_2=(5-1)/(2)=(4)/(2)=2 \end{gathered}`

so, we have 2 values for x ( 2 and 3)

Step 2

now, find the images

a) when x= 3

`\begin{gathered} f(x)=x^2-5x+3 \\ f(x)=(3)^2-5(3)+3 \\ f(x)=9-15+3 \\ f(x)=-3 \end{gathered}`

b) when x= 2

`\begin{gathered} f(x)=x^2-5x+3 \\ f(x)=2^2-5\cdot2+3 \\ f(x)=4-10+3 \\ f(x)=-3 \end{gathered}`

therefore the solutions are

`\begin{gathered} (2,-3)\text{ ;(3,-3)} \\ \end{gathered}`

I hope this helps you