Solve the equation answer in fractions.

3x+y=

Posted on Mar 10, 2009

The possible solutions are

x= -y/3

y= -3x/1

Posted on Feb 26, 2009

It seems to me that you are trying to solve the quadratic equation

aX^2+bX+c=10 with a=-3, b=3, c=15 or**-3X^2+3X+15=0**.

Since the all the coefficients are multiples of 3, one can simplify the equation by dividing every thing by 3, leaving -X^2+X+5=0. But to avoid confusing you I will consider the original equation**-3X^2+3X+15=0**..

You must first find out if the equation has any real solutions. To do that you calculate the discriminant (you do not have to remember the name if you choose to).

Discriminant is usually represented by the Greek letter DELTA (a triangle)

DELTA =b^2-4*a*c =(3)^2-4*(-3)*(15)=189

If the discriminant is positive (your case) the equation has two real solutions which are given by

**Solution1 =X_1=(-3-SQRT(189))/(-2*3)=(1+SQRT(21))/2**

**Solution2 =X_2=(-3+SQRT(189))/(-2*3)=(1-SQRT(21))/2** or about -1.791287847

Here SQRT stands for square root.

aX^2+bX+c=10 with a=-3, b=3, c=15 or

Since the all the coefficients are multiples of 3, one can simplify the equation by dividing every thing by 3, leaving -X^2+X+5=0. But to avoid confusing you I will consider the original equation

You must first find out if the equation has any real solutions. To do that you calculate the discriminant (you do not have to remember the name if you choose to).

Discriminant is usually represented by the Greek letter DELTA (a triangle)

DELTA =b^2-4*a*c =(3)^2-4*(-3)*(15)=189

If the discriminant is positive (your case) the equation has two real solutions which are given by

Here SQRT stands for square root.

Aug 17, 2014 | Educational & Reference Software

It seems to me that you are trying to solve the quadratic equation

aX^2+bX+c=10 with a=-3, b=3, c=15 or**-3X^2+3X+15=0**.

Since the all the coefficients are multiples of 3, one can simplify the equation by dividing every thing by 3, leaving -X^2+X+5=0. But to avoid confusing you I will consider the original equation**-3X^2+3X+15=0**..

You must first find out if the equation has any real solutions. To do that you calculate the discriminant (you do not have to remember the name if you choose to).

Discriminant is usually represented by the Greek letter DELTA (a triangle)

DELTA =b^2-4*a*c =(3)^2-4*(-3)*(15)=189

If the discriminant is positive (your case) the equation has two real solutions which are given by

**Solution1 =X_1=(-3-SQRT(189))/(-2*3)=(1+SQRT(21))/2**

**Solution2 =X_2=(-3+SQRT(189))/(-2*3)=(1-SQRT(21))/2** or about -1.791287847

Here SQRT stands for square root.

aX^2+bX+c=10 with a=-3, b=3, c=15 or

Since the all the coefficients are multiples of 3, one can simplify the equation by dividing every thing by 3, leaving -X^2+X+5=0. But to avoid confusing you I will consider the original equation

You must first find out if the equation has any real solutions. To do that you calculate the discriminant (you do not have to remember the name if you choose to).

Discriminant is usually represented by the Greek letter DELTA (a triangle)

DELTA =b^2-4*a*c =(3)^2-4*(-3)*(15)=189

If the discriminant is positive (your case) the equation has two real solutions which are given by

Here SQRT stands for square root.

Aug 17, 2014 | SoftMath Algebrator - Algebra Homework...

By definition, the y-intercept is the value of y when x=0. Just set x=0 in the equation and solve for y.

Hence 3(0)+6y=90 or 6y=90. This gives y=90/6=15. The point of intercept of the y-axis is (0,15).

To get the x-intercept, set y= 0 in the equation and solve for x.

3x+6(0)=90 or 3x=90. This gives x=90/3=30.

The point of intercept of the x-axis is (30,0).

Hence 3(0)+6y=90 or 6y=90. This gives y=90/6=15. The point of intercept of the y-axis is (0,15).

To get the x-intercept, set y= 0 in the equation and solve for x.

3x+6(0)=90 or 3x=90. This gives x=90/3=30.

The point of intercept of the x-axis is (30,0).

Jan 20, 2012 | SoftMath Algebrator - Algebra Homework...

4X+3-3=3X-4-3=3X-7

4X=3X-7

4X-3X=3X-3X-7

X=-7

The answer is X=-7

4X=3X-7

4X-3X=3X-3X-7

X=-7

The answer is X=-7

Dec 06, 2011 | Knowledge Adventure Math Blaster...

For the

3x-6(0)=-21. or 3x=-21 and x= -21/3=-7

Oct 31, 2011 | MathRescue Word Problems Of Algebra Lite

This equation doesn't have whole number factors so you have to solve it by completing the square or using the quadratic equation. I used the quadratic equation and got the following answers.

[3 + sqrt(209)]/10 and [3 - sqrt(209)]/10

I hope this helps you out.

[3 + sqrt(209)]/10 and [3 - sqrt(209)]/10

I hope this helps you out.

Jul 05, 2011 | SoftMath Algebrator - Algebra Homework...

x^2=-3x+28
x^2+3x - 28 = 0
D=9+112=121>0
x1=(-3+11)/2 = 4
x2=(-3-11)/2 = -7

Jun 03, 2010 | Bagatrix Precalculus Solved Full Version...

This is the process of how you would solve this.

-move the -3x over to the right hand side, so you have:

y = 5 + 3x

- since you know what x is, plug in the x which is 1:

y = 5 + 3(1)

- solve the equation

y = 5+3

y = 8, so the solution is 8.

-move the -3x over to the right hand side, so you have:

y = 5 + 3x

- since you know what x is, plug in the x which is 1:

y = 5 + 3(1)

- solve the equation

y = 5+3

y = 8, so the solution is 8.

Dec 16, 2008 | SoftMath Algebrator - Algebra Homework...

Hi joanmae jmeh;

You can add these equations just like you would numbers

3x + 2y = 7

5x - 2y = 1

-----------------------

8x =8

x=1

Now plug x = 1 back into either equation and you can solve for y

Y = 2

When you add the 2 equations together the +2y and the -2y cancel

out

Hope this helps Loringh Please leave a rating for me Thks

You can add these equations just like you would numbers

3x + 2y = 7

5x - 2y = 1

-----------------------

8x =8

x=1

Now plug x = 1 back into either equation and you can solve for y

Y = 2

When you add the 2 equations together the +2y and the -2y cancel

out

Hope this helps Loringh Please leave a rating for me Thks

Dec 01, 2008 | Bagatrix Algebra Solved! 2005 (105101) for...

(x+3y=0)*3=3x+9y

y=1 x= -3

y=1 x= -3

Jun 29, 2008 | Bagatrix Algebra Solved! 2005 (105101) for...

Nov 22, 2013 | Bagatrix Algebra Solved! 2005 (105101) for...

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