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4 years ago

Solve each system by substitution

Mathematics

1 answer:

Ira Lisetskai [31]4 years ago

4 0

Answer:

1. 2x + 3y =22

y=4

C. (5,4)

2. y=0

-5x - 2y = -15

B. (3,0)

Step-by-step explanation:

2x+(3)(4)=22

Step 1: Simplify both sides of the equation.

2x+12=22

Step 2: Subtract 12 from both sides.

2x+12−12=22−12

2x=10

Step 3: Divide both sides by 2.

2x/2 = 10/2

x=5

====================================================================

−5x−(2)(0)=−15

Step 1: Simplify both sides of the equation.

−5x−(2)(0)=−15

−5x+0=−15

(−5x)+(0)=−15(Combine Like Terms)

−5x=−15

Step 2: Divide both sides by -5.

−5x/−5 = −15/−5

x=3

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4 years ago

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Answer:

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4 years ago

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3 years ago

Which statement is true about the polynomial after it has been simplified. -10m^4n^3+8m^2n^6-6m^2n^6

laila [671]

Answer: 2m^2n^3 x ( -5m^2 + 4n^2 -3n^2 )

Step-by-step explanation:

ok, so just let me, uh, make this a little easier on my eyes lol:

-10m^4n^3+8m^2n^6-6m^2n^6 ----> (-10m^4)(n^3) + (8m^2)(n^6) - (6m^2)(n^6)

I just put in parentheses. nothing changed. just parentheses.

Since the leading coefficients, -10 , 8 , and 6 are multiples of two,

we can take out 2 from each... [-10/2 = 5 , 8/2 = 4 , 6/2 = 3 ]

= 2 x [ (-5m^4)(n^3) + (4m^2)(n^6) - (3m^2)(n^6) ]

Looking at it now...

we can take out m^2 from each... m^4 , m^2 , and m^2

m^2 [m^2 , 1 , 1]

since m^4 /m^2 = m^2 , m^2 /m^2 = 1 , m^2 /m^2 = 1

= 2(m^2) x [ (-5m^2)(n^3) + (4x1)(n^6) - (3x1)(n^6) ]

= 2(m^2) x [ (-5m^2)(n^3) + (4)(n^6) - (3)(n^6) ]

Finally, one more step...

we can take out n^3 from each term.... n^3 , n^6 , and n^6

n^3 [1 , n^2 , n^2] since n^3 /n^3 = 1 , n^6 /n^3 = n^2 , n^6 /n^3 = n^2

= 2(m^2)(n^3) [ (-5m^2)(1) + (4)(n^2) - (3)(n^2) ]

= 2m^2n^3 x [ (-5m^2) + (4n^2) - (3n^2) ]

= 2m^2n^3 x ( -5m^2 + 4n^2 -3n^2 )

ANSWER: 2m^2n^3 x ( -5m^2 + 4n^2 -3n^2 )

keep this parentheses. seriously.

7 0

3 years ago