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Dominik [7]
3 years ago
13

Ax+by=(a-b) , bx - ay =(a+b) simultaneous linear equations using cross multiplication

Mathematics
1 answer:
Molodets [167]3 years ago
6 0

Answer:

x=1\,,\,y=-1

Step-by-step explanation:

Given: ax+by=a-b\,,\,bx-ay=a+b

To solve: the given linear equations

Solution:

Consider the equations:

A_1x+B_1y+C_1=0\\A_2x+B_2y+C_2=0

By method of cross multiplication:

\frac{x}{B_1C_2-B_2C_1}=\frac{y}{C_1A_2-C_2A_1}=\frac{1}{A_1B_2-A_2B_1}

For equations: ax+by=a-b\,,\,bx-ay=a+b

ax+by-(a-b)=0\\bx-ay-(a+b)=0

Take A_1=a\,,\,B_1=b\,,\,C_1=-(a-b)\,,\,A_2=b\,,\,B_2=-a\,,\,C_2=-(a+b)

So,

\frac{x}{-b(a+b)-a(a-b)}=\frac{y}{-b(a-b)+a(a+b)}=\frac{1}{-a^2-b^2}\\\frac{x}{-ab-b^2-a^2+ab}=\frac{y}{-ab+b^2+a^2+ab}=\frac{1}{-(a^2+b^2)}\\\frac{x}{-(a^2+b^2)}=\frac{y}{a^2+b^2}=\frac{1}{-(a^2+b^2)}\\\frac{x}{-(a^2+b^2)}=\frac{1}{-(a^2+b^2)}\,,\,\frac{y}{a^2+b^2}=\frac{1}{-(a^2+b^2)}\\x=1\,,\,y=-1

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