Solve the equation answer in fractions.

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3x+y=

Posted on Mar 10, 2009

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The possible solutions are

x= -y/3

y= -3x/1

Posted on Feb 26, 2009

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Posted on Jan 02, 2017

5x (-3x) = 2x -3x - -3x = 0

The equation then becomes 2x=10 2x/2 = x and 10/2 =5

then x=5. does it work 5x5=10+3*5 YES

The equation then becomes 2x=10 2x/2 = x and 10/2 =5

then x=5. does it work 5x5=10+3*5 YES

Jan 20, 2017 | The Office Equipment & Supplies

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 | Computers & Internet

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...

For the

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

Oct 31, 2011 | MathRescue Word Problems Of Algebra Lite

Here you are :

2X2-3x+5=0 => 4-3x+5=0; (2X2=4) => -3x+9=0; (4+5=9) => 9 = 3x; (-3x moved to the right-hand side and it became 3x) => 9/3=x; (3 moved to the right hand side, so it becomes 1/3, i.e 1/3X9= 9/3) i.e. x=9/3 => x=3

Hence x equals to 3.

You can add or remove the explanation/description, I simply add them to make you understand.

If you need help, let me know.

Good luck.

Thanks for using** FixYa.**

2X2-3x+5=0 => 4-3x+5=0; (2X2=4) => -3x+9=0; (4+5=9) => 9 = 3x; (-3x moved to the right-hand side and it became 3x) => 9/3=x; (3 moved to the right hand side, so it becomes 1/3, i.e 1/3X9= 9/3) i.e. x=9/3 => x=3

Hence x equals to 3.

You can add or remove the explanation/description, I simply add them to make you understand.

If you need help, let me know.

Good luck.

Thanks for using

Oct 08, 2010 | Computers & Internet

Hello,

Sorry, but what you wrote is not an equation but a polynomial expression. You want to solve the equation x^4+5x^3-3x^2-43x-60 =0.

The solve( command, can only be used with real numbers.

The** solve(** is available through the CATALOG :
[2nd][CATALOG], scroll down till you reach the command. Highlight it
and press [ENTER]. The command echodes on main screen as **solve(** .

You complete the command by entering the expression (not the equation), the name of the variable you solve for, the initial guess , and { lower limit, upper limit} between curly brackets, and the closing parenthesis.

Exemple:

**solve (x^4+5x^3-3x^2-43x-60 ****, x,0 {-5,0} ) [ENTER]**

should give you the negative root,

**solve (x^4+5x^3-3x^2-43x-60 ****, x,0 {0,5} ) [ENTER]**

should give you the positive root.

It is implied that the expression is 0, so you should not insert =0, otherwise you get an error. Here for the lower limit is -5 you must use the change sign symbol (-) under the 3 key, not the regular MINUS.

You may ask how I knew that there were two roots when the equation is a quartic? By first graphing it to have an idea about where the roots lie and how many there are. You should always do that to speed up the search.

There is another way to zoom in on the roots: by drawing the graph and using the tools accessible under the [2nd][CALC] menu, namely the option [2:Zero]

The resolution of the TI83/84 is not good enough for this function that grows too fast, but I am inserting a picture of the curve from another calculator with a much better resolution.

Hope it helps.

Sorry, but what you wrote is not an equation but a polynomial expression. You want to solve the equation x^4+5x^3-3x^2-43x-60 =0.

The solve( command, can only be used with real numbers.

The

You complete the command by entering the expression (not the equation), the name of the variable you solve for, the initial guess , and { lower limit, upper limit} between curly brackets, and the closing parenthesis.

Exemple:

should give you the negative root,

should give you the positive root.

It is implied that the expression is 0, so you should not insert =0, otherwise you get an error. Here for the lower limit is -5 you must use the change sign symbol (-) under the 3 key, not the regular MINUS.

You may ask how I knew that there were two roots when the equation is a quartic? By first graphing it to have an idea about where the roots lie and how many there are. You should always do that to speed up the search.

There is another way to zoom in on the roots: by drawing the graph and using the tools accessible under the [2nd][CALC] menu, namely the option [2:Zero]

The resolution of the TI83/84 is not good enough for this function that grows too fast, but I am inserting a picture of the curve from another calculator with a much better resolution.

Hope it helps.

Oct 18, 2009 | Texas Instruments TI-83 Plus Calculator

Just plug in one side of the equation and find the answer. Then plug in the other side and compare answers.

Feb 19, 2008 | Texas Instruments TI-89 Calculator

A positive number is 5 times another number.If 21 is added to both the numbers,then one of the new numbers becomes twice the other number.What are the numbers

Nov 29, 2007 | Computers & Internet

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