Solve {l}{y=x^2-4x+3}{x=15/4}

$$\left. \begin{array} { l } { y = x ^ { 2 } - 4 x + 3 } \\ { x = \frac { 15 } { 4 } } \end{array} \right.$$

Solve for y, x

$$x = \frac{15}{4} = 3\frac{3}{4} = 3.75$$

$$y = \frac{33}{16} = 2\frac{1}{16} = 2.0625$$

Solution Steps

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

$$y=\left(\frac{15}{4}\right)^{2}-4\times \frac{15}{4}+3$$

Calculate \frac{15}{4} to the power of 2 and get \frac{225}{16}.

$$y=\frac{225}{16}-4\times \frac{15}{4}+3$$

Multiply -4 and \frac{15}{4} to get -15.

$$y=\frac{225}{16}-15+3$$

Subtract 15 from \frac{225}{16} to get -\frac{15}{16}.

$$y=-\frac{15}{16}+3$$

Add -\frac{15}{16} and 3 to get \frac{33}{16}.

$$y=\frac{33}{16}$$

The system is now solved.

$$y=\frac{33}{16} x=\frac{15}{4}$$