Solve {l}{f{(x)}=px^2-{(p+3)}x+4}{x=1}{g=1}{text{Solvefor}htext{where}}{h=g}

$$\left. \begin{array} { l } { f {(x)} = p x ^ {2} - {(p + 3)} x + 4 }\\ { x = 1 }\\ { g = 1 }\\ { \text{Solve for } h \text{ where} } \\ { h = g } \end{array} \right.$$

Solve for x

$$x=1$$

$$f=1\text{ and }g=1\text{ and }h=1$$

Solution Steps

Consider the first equation. Insert the known values of variables into the equation.

$$f\times 1=p\times 1^{2}-\left(p+3\right)\times 1+4$$

Calculate 1 to the power of 2 and get 1.

$$f\times 1=p\times 1-\left(p+3\right)\times 1+4$$

Use the distributive property to multiply p+3 by 1.

$$f\times 1=p\times 1-\left(p+3\right)+4$$

To find the opposite of p+3, find the opposite of each term.

$$f\times 1=p\times 1-p-3+4$$

Combine p\times 1 and -p to get 0.

$$f\times 1=-3+4$$

Add -3 and 4 to get 1.

$$f\times 1=1$$

Divide both sides by 1.

$$f=1$$

The system is now solved.

$$f=1 x=1 g=1 h=1$$