Solve the system by finding the determinants and using Cramer's Rule: 2x - y = 4 3x + y = 1

1 answer:

Ahat [919]3 years ago

5 0

Answer:

(1, - 2)

Step-by-step explanation:

2x - y = 4

3x + y = 1

A = $\left[\begin{array}{cc}2&-1\\3&1\end{array}\right]$ = 2(1) - 3( - 1) =2 + 3 = 5

$A_{x}$ = $\left[\begin{array}{cc}4&-1\\1&1\end{array}\right]$ = 4(1) - 1(- 1) = 4 + 1 = 5

$A_{y}$ = $\left[\begin{array}{cc}2&4\\3&1\end{array}\right]$ = 2(1) - 4(3) = 2 - 12 = - 10

x = $\frac{A_{x} }{A}$ = 1

y = $\frac{A_{y} }{A}$ = - 2

(1, - 2)

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Conversion of a quadratic equation from standard form to vertex form is done by completing the square method.

Assume the quadratic equation to be $\mathbf{ax^{2}+bx+c=0}$ where x is the variable.

Completing the square method is as follows:

send the constant term to other side of equal $\mathbf{ax^{2}+bx=-c}$
divide the whole equation be coefficient of $\mathbf{x^{2}}$ , this will give $\mathbf{x^{2}+\frac{b}{a}x=- \frac{c}{a}}$
add $\mathbf{(\frac{b}{2a})^{2}}$ to both side of equality $\mathbf{x^{2}+2\times\frac{b}{2a}x+\frac{b}{2a}^{2}=-\frac{c}{a}+\frac{b}{2a}^{2}}$
Make one fraction on the right side and compress the expression on the left side $\mathbf{(x+\frac{b}{2a})^{2}=\frac{b^{2}-4ac}{4a^{2}}}$
rearrange the terms will give the vertex form of standard quadratic equation $\mathbf{a(x+\frac{b}{2a})^{2}-\frac{b^{2}-4ac}{4a}=0}$