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Stels [109]
2 years ago
9

A student is trying to solve the system of two equations given below: equation p: y z = 6 equation q: 5y 9z = 1 which of these i

s a possible step used in eliminating the y-term? (y z = 6) ⋅ 9 (y z = 6) ⋅ −5 (5y 9z = 1) ⋅ 9 (5y 9z = 1) ⋅ 5
Mathematics
1 answer:
KiRa [710]2 years ago
7 0

System of equation consists of two or more equations with unknow variables. A possible step used in eliminating the y-term is (y + z  6) *5

<h3>System of equations</h3>

System of equation consists of two or more equatons with unknow variables.


Given the system of equations

y + z = 6

5y + 9z =1

In order to eliminate the variable "y", we <em>will multiply equation 1 by 5 and subtract from equation 2;</em>

5y + 5z = 30

5y + 9z = 1

Subtract

5z - 9z = 30 - 1

-4z =29

This leaves only the variable "z" eliminating variable. "y"

Learn more on system of equation here: brainly.com/question/14323743

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Step-by-step explanation:

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2 years ago
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True or False.<br> The expressions 3 + y and 1/3 (6 + y) are equivalent?
vazorg [7]

Answer:

false

Step-by-step explanation:

consider y = 3

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3 years ago
100 POINTS PLS HELP write the value of six dimes as a fracition,with 100 as a demominator
BartSMP [9]

1 dimes are worth 10 cents.

Multiply.

6 * 10 = 60

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Best of Luck!

7 0
3 years ago
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Find number a and k so that x-2 is afactor<br>OF F(x)=x4-2ax tax-<br>X+k and F(-1)=3​
Slav-nsk [51]

Question:

Find numbers a and k so that x-2 is a factor of

f(x)=x^4-2ax^3+ax^2- x+k

and

f(-1)=3

Answer:

k = -2 and a=1

Step-by-step explanation:

Given

f(x)=x^4-2ax^3+ax^2- x+k

Factor:\ x - 4

f(-1)=3

Required

Find a and k

For f(-1)=3

Substitute -1 for x

f(x)=x^4-2ax^3+ax^2- x+k

f(-1) = (-1)^4 - 2a *(-1)^3 + a*(-1)^2 - (-1) + k

f(-1) = 1 - 2a *-1 + a*1 +1 + k

f(-1) = 1 +2a + a +1 + k

f(-1) = 2 +3a + k

Substitute 3 for f(-1)

3 = 2 +3a + k

Collect Like Terms

3 - 2 = 3a + k

1 = 3a + k

Also:

If x - 2 is a factor, then

f(2) = 0

Substitute 2 for x and 0 for f(x)

f(x)=x^4-2ax^3+ax^2- x+k

0 = 2^4 - 2a * 2^3 + a * 2^2 - 2 + k

0 = 16 - 2a * 8 + a * 4 - 2 + k

0 = 16 - 16a + 4a - 2 + k

0 = 16 - 12a - 2 + k

Collect Like Terms

2 - 16 = - 12a + k

-14 = k - 12a

k = 12a - 14

Substitute 12a - 14 for k in 1 = 3a + k

1 = 3a + 12a - 14

1 = 15a - 14

Collect Like Terms

15a = 1 + 14

15a = 15

Solve for a

a = \frac{15}{15}

a=1

Substitute 1 for a in k = 12a - 14

k = 12 * 1 - 14

k = 12 - 14

k = -2

3 0
2 years ago
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