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4 years ago

-9x+10y = -18 3x- y =6 Elimination

2 answers:

8 0

Answer: x = 2 and y = 0

Explanation:

-9x + 10y = -18
3x - y = 6
————————
-9x + 10y = -18
10(3x - y = 6)
————————
-9x + 10y = -18
30x - 10y = 60
————————-
21x = 42
x = 42/21
x = 2

Find y:

3x - y = 6
3(2) - y = 6
6 - y = 6
-y = 6 - 6
-y = 0
y = 0

Fiesta28 [93]4 years ago

3 0

Step-by-step explanation:

Hey, there!!

To solve this type of equation it is very easy from elimination method,

Firsty, you try to search if there is opposite sign of same variables or not in both equation.
Then find their LCM to make easy to cancel.
And thirdly add or subtract the terms to get value of one variable.
Then substitute the value of one variable to get value of another variable.

Here, the given equation are,

-9x+10y = -18..........(i)

3x-y = 6....................(ii)

Let's take LCM to solve for x.

To solve for "x" we must cancel "y".

LCM= 10 { lcm of 1 and 10 is 1}

So, let's multiply equation (ii) by 10 and add them.

Then,

-9x + 10y = -18

30x - 10y = 60

21x = 42 { as +10y and -10y gets cancelled}

$x = \frac{42}{21}$

Therefore, x= 2.

Substituting the value of x in equation (i)

-9x + 10y = -18

-9 × 2 +10y = -18

-18 + 10y = -18

or, 10y = -18 +18

Therefore, y= 0.

So, the value of x is 2 and y is 0.

Hope it helps...

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Which expressions are equivalent to 35+30s-45t35+30s−45t35, plus, 30, s, minus, 45, t? Choose 2 answers:

laila [671]

Answer:

Options B

Options C

Options D

Step-by-step explanation:

35 + 30s - 45t

First term, 35

Factors:

= 1, 5, 7

Second term, 30s

Factors:

= 1, 2, 3 ,5

Third term, 45t

Factors:

= 1, 3, 3, 5

Common factors = 1, 5

5 × (7 + 6s - 9t)

10 × (3.5 + 3s - 4.5t)

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3 years ago

Find the sum of -3x2 -4x+3 and 2x2-x+3

Alika [10]

Answer:

Assuming the 2 at the ends are the exponents,

the answer is -x^2-5x+6 (x^2+5x-6)

from here, it cannot be simplified more

Explanation --

Just add everything by putting the terms in parentheses and adding them.

Keep simplifying by putting the like terms together and adding/subtracting them.

At the end, you come up with -x^2-5x+6, so multiply -1 on the whole thing (by putting parentheses on its sides. This makes x positive. But they are the same thing, but this step is just for simplifying the end product a little more.

Hope this helps!! Have a nice day!

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3 years ago

What is the first step to solving the following equation?5x –11 = 42

alexdok [17]

Answer:

add the 11 to both sides to cancel it out.

Step-by-step explanation:

so basically

5x - 11 + 11 = 42 +11

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3 years ago

Find 120% of 18. a) 22.5 b) 23 c) 15 d) 21.6

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120% of 18 is =


D) 21.6

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3 years ago

Use the definition of continuity to determine whether f is continuous at a.

dmitriy555 [2]

$f(x)$ will be continuous at $x=a=7$ if
(i) $\displaystyle\lim_{x\to7}f(x)$ exists,
(ii) $f(7)$ exists, and
(iii) $\displaystyle\lim_{x\to7}f(x)=f(7)$ .

The second condition is immediate, since $f(7)=8918$ has a finite value. The other two conditions can be established by proving that the limit of the function as $x\to7$ is indeed the value of $f(7)$ . That is, we must prove that for any $\varepsilon>0$ , we can find $\delta>0$ such that

$|x-7|$

Now,

$|f(x)-f(7)|=|5x^4-9x^3+x-8925|$

Notice that when $x=7$ , we have $5x^4-9x^3+x-8925=0$ . By the polynomial remainder theorem, we know that $x-7$ is then a factor of this polynomial. Indeed, we can write

$|5x^4-9x^3+x-8925|=|(x-7)(5x^3+26x^2+182x+1275)|=|x-7||5x^3+26x^2+182x+1275|$

This is the quantity that we do not want exceeding $\varepsilon$ . Suppose we focus our attention on small values $\delta$ . For instance, say we restrict $\delta$ to be no larger than 1, i.e. $\delta\le1$ . Under this condition, we have

$|x-7|$

Now, by the triangle inequality,

$|5x^3+26x^2+182x+1275|\le|5x^3|+|26x^2|+|182x|+|1275|=5|x|^3+26|x|^2+182|x|+1275$

If $|x|$ , then this quantity is moreover bounded such that

$|5x^3+26x^2+182x+1275|\le5\cdot8^3+26\cdot8^2+182\cdot8+1275=6955$

To recap, fixing $\delta\le1$ would force $|x|$ , which makes

$|x-7||5x^3+26x^2+182x+1275|$

and we want this quantity to be smaller than $\varepsilon$ , so

$6955|x-7|$

which suggests that we could set $\delta=\dfrac{\varepsilon}{6955}$ . But if $\varepsilon$ is given such that the above inequality fails for $\delta=\dfrac{\varepsilon}{6955}$ , then we can always fall back on $\delta=1$ , for which we know the inequality will hold. Therefore, we should ultimately choose the smaller of the two, i.e. set $\delta=\min\left\{1,\dfrac{\varepsilon}{6955}\right\}$ .

You would just need to formalize this proof to complete it, but you have all the groundwork laid out above. At any rate, you would end up proving the limit above, and ultimately establish that $f(x)$ is indeed continuous at $x=7$ .

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