Question

Solve each of the following systems of equations by the method of cross-multiplication:
2(ax-by)+a+4b=02(bx+ay)+b-4a=0

Answer 1

From the solved given equations by using Cross multiplication method we get the values of x and y are $\frac{-1}{2}$ and 2 respectively.

Step-by-step explanation:

Given equations are $2(ax-by)+a+4b=0\hfill (1)$ and $2(bx+ay)+b-4a=0\hfill (2)$

To solve the given equations by using the method Cross Multiplication :

From equation (1) we have

• $2(ax-by)+a+4b=0$
• $2ax-2by+a+4b=0\hfill (3)$

From equation (2) we have

• $2(bx+ay)+b-4a=0$
• $2bx+2ay+b-4a=0\hfill (4)$

The above two equations (3) and (4) is of the form

$a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$

We have the formula $\frac{x}{b_1c_2-b_2c_1}=\frac{y}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$

• Now solving the equations (3) and (4) we get
• By Cross Multiplication method we have

     x            y         1

-2b    a+4b     2a      -2b

2a      b-4a     2b      2a

By using the formula

• $\frac{x}{-2b)(b-4a)-2a(a+4b)}=\frac{y}{2b(a+4b)-2a(b-4a)}=\frac{1}{4a^2+4b^2}$

Now equating  x and constant

• $\frac{x}{-2b^2+8ba-2a^2-8ab}=\frac{1}{4a^2+4b^2}$
• $\frac{x}{-2a^2-2b^2}=\frac{1}{4a^2+4b^2}$
• $x=\frac{1(-2a^2-2b^2)}{4a^2+4b^2}$
• $x=\frac{-2(a^2+b^2)}{4(a^2+b^2)}$

• $x=\frac{-1}{2}$

Therefore the value of x is $\frac{-1}{2}$

Now equating y and constant

• $\frac{y}{2b(a+4b)-2a(b-4a)}=\frac{1}{4a^2+4b^2}$
• $\frac{y}{2ba+8b^2-2ab+8a^2}=\frac{1}{4a^2+4b^2}$
• $\frac{y}{8b^2+8a^2}=\frac{1}{4a^2+4b^2}$
• $\frac{y}{8(a^2+b^2)}=\frac{1}{4(a^2+b^2)}$
• $y=\frac{8(a^2+b^2)}{4(a^2+b^2)}$
• $y=2$

Therefore the value of y is 2

From the solved given equations we have the values of x and y are $\frac{-1}{2}$ and 2 respectively