3 years ago

Don't know what to do

Mathematics

1 answer:

aivan3 [116]3 years ago

3 0

Graph those on the number line

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

Molodets [167]

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,

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

6 0

3 years ago

Simplify the expression 5(n^2+n)-3n(2n^2+4n-2)

slega [8]

first, lets distribute the equation. 5(n^2+n)-3n(2n^2+4n-2) would become

5n^2 +5n -6n^3 -12n^2 +6n

Now we add the like terms.

-6n^3 - 7n^2 +11n

8 0

3 years ago

Solve: <br> 10c + 20 = 40

Natali5045456 [20]

Subtract 20 from 40 then divide by 10

C=2

8 0

3 years ago

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(60y – 18) = 2(8-16)

Alekssandra [29.7K]

Answer:

3/10

Step-by-step explanation:

(60y-18)=2(8-16)

60y-18=16-16

60y-18=0

60y=18 /:60

y= 18/60=3/10

4 0

3 years ago

PLEASE PLEASE HELP ME, NO LINKS, I JUST NEED THIS DONE

lianna [129]

Answer:

Here is your answer

Step-by-step explanation:

B) 2x + 5 = 7

vertical line

C) 2x - y = 6

oblique line

D) 3y + 9 = 0

horizontal line

6 0

3 years ago

Read 2 more answers