What is the value of y in the following system y= 3x-5 6x+3y=15

2 answers:

7 0

X=2. You use substitution to put the y=3x-5 into 6x+3y=15.
So you get 6x+3(3x-5)=15
=6x+9x-15=15
add 15 on both sides to get the x's alone and add the x's together.
6+9=15 -- 15+15=30
15x=30
30/15 =2
x=2

Ivahew [28]2 years ago

3 0

If you plug in (3x-5) in for y in the 6x+3y=15, you will get 6x+9x-15=15. Add the like terms (6x+9x) and get 15x. Bring the -15=15 down. Add 15 to the other 15 on the other side of the equal sign, bring 15x down and then divide 15 by 30 and get x=2. Plug in 2 for x in the (3x-5) equation, then solve and the answer will be 1.

You might be interested in

ellen sells mugs that are in boxes in the shape of a cube.The edges of the boxes have a lenght of 1/2 foot.Ellen can completely

Elanso [62]

Answer: 189^2

Step-by-step explanation:

5 0

1 year ago

A plumbing contractor bid the rough-in work for a new three-story office building based an estimate of 850 hours of a journey

Elodia [21]

There is 179 hours left.

-Hiadamcom

6 0

1 year ago

What is the quotient (3x4 – 4x2 + 8x – 1) ÷ (x – 2) brainly

Karolina [17]

To find our quotient we are going to perform long division.

Long division step by step:

Step 1. Write the polynomial (dividend) in descending order (from highest to least exponent). If you encounter a missing term, use zero to fill the space of the term in the descending sequence.

Notice that even tough our polynomial is written in descending form, $x^3$ is missing, so we are going to use $0x^3$ to fill the gap:

$3x^4-4x^2+8x-1$

$3x^4+0x^3-4x^2+8x-1$

Step 2. Divide the first term of the polynomial (dividend) by the first term of the divisor.

In our case the first term of the dividend is $3x^4$ and the first term of the divisor is $x$ , so $\frac{3x^4}{x} =3x^3$ . Notice that $3x^3$ is the first term of the quotient.

Step 3. Multiply the first term of the quotient by the terms of the divisor and subtract them from the respective term of the dividend.

The first term of our quotient (from the previous step) is $3x^3$ and the divisor is $(x-1)$ , so $3x^3(x-2)=3x^4-6x^3$ . Now, we are going to subtract them from the respective term (the term with the same power) of the dividend. The respective terms of the dividend are $3x^4$ and $0x^3$ , so $3x^4-3x^4=0$ and $0x^3-(-6x^3)=0x^3+6x^3=6x^3$

Step 4. Bring down the next term in the dividend and repeat the process for the remaining terms.

After finish the process (check the attached picture), we can conclude that the quotient of $3x^4-4x^2+8x-1$ ÷ $(x-2)$ is $3x^3+6x^2+8x+24$ with a remainder of $47$ , or in a different notation: $3x^3+6x^2+8x+24+\frac{47}{x-2}$