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2 years ago

Solve each system by elimination. 1) 10x + 7y=9 -4x - 7y=9

Mathematics

1 answer:

stellarik [79]2 years ago

8 0

Answer:

(3, -3)

Step-by-step explanation:

When asked to solve by elimination, you put them on top of one another, like you're going to add it.

10x + 7y = 9

-4x - 7y = 9

See that 7y? You can cancel those out because one is negative, and one is positive. So those are gone. You finish adding the rest of the numbers as usual and solve for x.

6x = 18

x = 3

Take x, and plug it into either equation to find y.

10(3) + 7y = 9

7y = -21

y = -3

(3, -3)

Hope this helped!

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How to solve 5h+7=17

Alex777 [14]

If you are trying to find h, then its very simple.
So lets see 5+7=12
So 17-12=5
So lets check to see if h=5
5+5+7
5+5=10+7=17 so h is 5. :)

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3 years ago

Renee adds 5 to a number, then multiplies the sum by-2. The result is 8. Write and solve an equation to find the number, x. What

miskamm [114]

5+(-9)= -4 • -2 =8
i hope i’m right.

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3 years ago

I need help with 2 and 5 and I need you to right down how you did it plz and thank you

anastassius [24]

Number 5 is 22 and 35
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2 years ago

In 2000 the total amount of gamma ray bursts was recorded at 6.4 million for a city. In 2005, the same survey was made and the t

ratelena [41]

Let's assume

It started in 2000

so, t=0 in 2000

$P_0=6.4million$

we can use formula

$P(t)=P_0 e^{rt}$

we can plug value

$P(t)=6.4 e^{rt}$

In 2005, the same survey was made and the total amount of gamma ray bursts was 7.3 million

so, at t=2005-2000=5

P(t)=7.3 million

we can plug value and then we can solve for r

$7.3=6.4 e^{r*5}$

$r=0.02632$

now, we can plug back

$P(t)=6.4 e^{0.02632t}$

now, we have

P(t)=1 billion =1000 million

so, we can set it and then we can solve for t

$1000=6.4 e^{0.02632t}$

$t=191.924$

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Year is 2000+192

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4 0

3 years ago

a collection of dimes and quarters is worth $19.85. There are 128 coins in all. How many of each type of coin are in the collect

PolarNik [594]

Number of dimes were 81 and number of quarters were 47

Solution:

Let "d" be the number of dimes

Let "q" be the number of quarters

We know that,

value of 1 dime = $ 0.10

value of 1 quarter = $ 0.25

Given that There are 128 coins in all

number of dimes + number of quarters = 128

d + q = 128 ------ eqn 1

Also given that collection of dimes and quarters is worth $19.85

number of dimes x value of 1 dime + number of quarters x value of 1 quarter = 19.85

$d \times 0.10 + q \times 0.25 = 19.85$

0.1d + 0.25q = 19.85 -------- eqn 2

Let us solve eqn 1 and eqn 2

From eqn 1,

d = 128 - q -------- eqn 3

Substitute eqn 3 in eqn 2

0.1(128 - q) + 0.25q = 19.85

12.8 - 0.1q + 0.25q = 19.85

12.8 + 0.15q = 19.85

0.15q = 7.05

Therefore from eqn 3,

d = 128 - q

d = 128 - 47

Thus number of dimes were 81 and number of quarters were 47

4 0

3 years ago