All categories

3 years ago

Is (8, 3) a solution to the system? - x + 4y = 4 - x + 3y = 1

Mathematics

1 answer:

Kaylis [27]3 years ago

8 0

Answer:

Step-by-step explanation:

You can either substitute (8, 3) into both equations as a test, or you can find the solution yourself.

Rewrite

- x + 4y = 4

- x + 3y = 1

- x + 4y = 4

+ x - 3y = -1

and combine like terms. The result is y = 3. Then -x + 4(3) = 4, or 8 = x.

Yes, (8, 3) is a solution of this SYSTEM.

You might be interested in

|-4/9|+5/6 write answer in simplest form

Mazyrski [523]

Answer: 23/18

Step-by-step explanation:

8 0

3 years ago

Find the exact solution of the system of equations of x^2+y^2-4x-6y+4=0 and x^2-4x-3y+4=0

gavmur [86]

We can readily know the x^2-4x+4=3y then use it to replace the same function in the first equation which refers to the 3y+y^2-6y=0
y^2-3y=0
y(y-3)=0
y1=0 -----------y2=3

7 0

4 years ago

Harvey the wonder hamster can run 3\dfrac16 \text{ km}3 6 1 km3, start fraction, 1, divided by, 6, end fraction, start text, s

Snowcat [4.5K]

Answer:

$Speed = \frac{38}{3}km/hr$

Step-by-step explanation:

Given

$Distance = 3\frac{1}{6}km$

$Time = \frac{1}{4}hr$

Required

Determine the average speed

Average speed is calculated as thus;

$Speed = \frac{Distance}{Time}$

$Speed = 3\frac{1}{6}/\frac{1}{4}$

$Speed = 3\frac{1}{6} * \frac{4}{1}$

Convert Mixed Fraction to Improper fraction

$Speed = \frac{19}{6} * \frac{4}{1}$

$Speed = \frac{19 * 4}{6 * 1}$

$Speed = \frac{76}{6}$

Divide both sides of the fraction by 2

$Speed = \frac{38}{3}$

Hence:

The speed is:

$Speed = \frac{38}{3}km/hr$

3 0

3 years ago

Solve for x and y x-y=11 2x+y=19

Ilia_Sergeevich [38]

X=10 y=-1
U can use Photomath for questions like this

7 0

3 years ago

Read 2 more answers

How do I solve this word problem?

german

Answer:

280 student tickets
520 adult tickets

Step-by-step explanation:

You may recognize that you are given two relationships between two unknowns. You can write equations for that.

You are asked for numbers of adult tickets and of student tickets. It often works well to let the values you're asked for be represented by variables. We can choose "a" for the number of adult tickets, and "s" for the number of student tickets. Then the problem statement tells us the relationships ...

a + s = 800 . . . . . . 800 tickets were sold

12.50a + 7.50s = 8600 . . . . . . . revenue from sales was 8600

(You are supposed to know that the revenue from selling "a" adult tickets is found by multiplying the ticket price by the number of tickets: 12.50a.)

___

You can solve these two equations any number of ways. One way is to do it by elimination. We can multiply the first equation by 12.50 and subtract the second equation:

12.50(a +s) -(12.50a +7.50s) = 12.50(800) -(8600)

5s = 1400 . . . . simplify. (The "a" variable has been eliminated.)

s = 280 . . . . . . divide by 5

Then the number of adult tickets can be found from the first equation:

a + 280 = 800

a = 520

280 student tickets and 520 adult tickets were sold.

8 0

4 years ago