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4 years ago

Use elimination to solve each system of equations.

Mathematics

1 answer:

Allushta [10]4 years ago

7 0

Answer:

x=-3

Y=-5

Step-by-step explanation:

2x-y=-1 ×2

3x-2y=1. ×1

4x-2y= -2

3x-2y=1

eliminate y by subtracting

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Solve for y 1.2x - 4y = 28

Dovator [93]

Your question is incomplete i guess

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4 years ago

Simplify (a+b)^3 + (a-b)^3 + 6a(a^2-b^2)

11111nata11111 [884]

Answer:

8a^3.

Step-by-step explanation:

(a+b)^3=a^3+b^3+3a^2b+3ab^2

(a-b)^3=a^3-b^3-3a^2b+3ab^2

(a+b)^3+(a-b)^3=2a^3+6ab^2

According to the question

(a+b)^3+(a-b)^3+6a(a^2-b^2)

Put in the value

=2a^3+6ab^2 +6a^3–6ab^2

=8a^3

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4 years ago

What is another way to express 42 +24?

-BARSIC- [3]

Answer:

42*2

Step-by-step explanation:

42*2

8 0

3 years ago

Read 2 more answers

PLEASEE SOMEONE ANSWER ILL DO ANYTHING ANYTHING PLEAASEEE HELP ME this is myh 6th time posting this ;[

Sonbull [250]

What is the problem?

4 0

3 years ago

What is the correct first step to solve this system of equations by elimination?

storchak [24]

$\bold{\huge{\pink{\underline{ Solution }}}}$

$\bold{\underline{ Given }}$

We have given two linear equations that is 2x - 3y = -6 and x + 3y = 12 .

$\bold{\underline{ To \: Find }}$

We have to find the value of x and y by elimination method.

$\bold{\underline{ Let's \: Begin }}$

$\sf{ 2x - 3y = -6 ...eq(1)}$

$\sf{ x + 3y = 12 ...eq(2)}$

Multiply eq( 2 ) by 2 :-

$\sf{ 2( x + 3y = 12 )}$

$\sf{ 2x + 6y = 24 }$

Subtract eq(1) from eq(2) :-

$\sf{ 2x + 6y -( 2x - 3y) = 24 -(-6)}$

$\sf{ 2x + 6y - 2x + 3y = 24 + 6 }$

$\sf{ 9y = 30 }$

$\sf{ y = 30/9}$

$\sf{\red{ y = 10/3}}$

Now, Subsitute the value of y in eq( 1 ):-

$\sf{ 2x - 3(10/3) = -6 }$

$\sf{ 2x - 10 = -6 }$

$\sf{ 2x = -6 + 10}$

$\sf{ x = 4/2}$

$\sf{\red{ x = 2}}$

Hence, The value of x and y is 2 and 10/3

6 0

3 years ago