Question about Bagatrix Algebra Solved! 2005 (105101) for PC

3x+2y= 7

5x-2y= 1 is my problem.

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

Posted on Dec 01, 2008

Solve 4x-2y=8

Posted on Mar 02, 2009

(0,2)

Posted on Dec 31, 2008

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Posted on Jan 02, 2017

First, we will find y in terms of x. We will use the first equation to determine this.

4x+2y=2

We can subtract 4x from both sides:

2y=2-4x

And then divide both sides of the equation by two:

y=1-2x

Since we now have y in terms of x, we can substitute this into our second equation.

-3x-y=-3

-3x-(1-2x)=-3

Then, we can distribute the minus sign

-3x-1+2x=-3

-x-1=-3

Next, we can add 1 to both sides of the equation.

-x=-2

Finally, we divide both sides by negative one to isolate x.

x=2

Now that we have x's value, we can find y's value.

The first thing that we determined is:

y=1-2x

We can substitute in the value of x to this equation.

y=1-2x

y=1-4

y=-3

Therefore, we now have the values of both variables.

x=2

y=-3

4x+2y=2

We can subtract 4x from both sides:

2y=2-4x

And then divide both sides of the equation by two:

y=1-2x

Since we now have y in terms of x, we can substitute this into our second equation.

-3x-y=-3

-3x-(1-2x)=-3

Then, we can distribute the minus sign

-3x-1+2x=-3

-x-1=-3

Next, we can add 1 to both sides of the equation.

-x=-2

Finally, we divide both sides by negative one to isolate x.

x=2

Now that we have x's value, we can find y's value.

The first thing that we determined is:

y=1-2x

We can substitute in the value of x to this equation.

y=1-2x

y=1-4

y=-3

Therefore, we now have the values of both variables.

x=2

y=-3

Jan 13, 2015 | SoftMath Algebrator - Algebra Homework...

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...

Try casting the equation of the circle into the form (x-h)^2+(y-k)^2=R^2. Here (h,k) would be the center and R the radius.

To do that complete the two squares.

x^2-2x=x^2-2x+1-1=(x^2-2x+1)-1=(x-1)^2 -1.

Do the same for the y terms.

y^2-2y=(y-1)^2 -1.

Substitute these two expressions in your original equation.

(x-1)^2+(y-1)^2 -1-1=14 or (x-1)^2+(y-1)^2=14+2=16

The radius is square root of 16 or 4, and the center is C(1,1).

To do that complete the two squares.

x^2-2x=x^2-2x+1-1=(x^2-2x+1)-1=(x-1)^2 -1.

Do the same for the y terms.

y^2-2y=(y-1)^2 -1.

Substitute these two expressions in your original equation.

(x-1)^2+(y-1)^2 -1-1=14 or (x-1)^2+(y-1)^2=14+2=16

The radius is square root of 16 or 4, and the center is C(1,1).

Sep 20, 2011 | SoftMath Algebrator - Algebra Homework...

Allan is 19 and Roy is 13.

Solution:

First we have 2 variables, their ages: I used Allan = x & Roy = y

Equation 1: X - 4 = 2*(Y-4) - 3

We can ignore the 5 years in the second part of the statement. Allan will always be 6 years older than Roy whether it be 5 years from now or ten years form now.

Equation 2: X = Y + 6

Substitute Y + 6 for X in equation 1:

X - 4 = 2*(Y-4) - 3 becomes Y + 6 - 4 = 2*(Y-4) - 3

That simplifies to Y + 2 = 2Y - 8 - 3

Rewrite it again: Y + 2 = 2Y - 11

Add 11 to both sides: Y + 13 = 2Y

Subtract Y from both sides: 13 = Y

Now plug 13 in for y in Equation 2

X = Y + 6

X = 13 + 6

X = 19

Solution:

First we have 2 variables, their ages: I used Allan = x & Roy = y

Equation 1: X - 4 = 2*(Y-4) - 3

We can ignore the 5 years in the second part of the statement. Allan will always be 6 years older than Roy whether it be 5 years from now or ten years form now.

Equation 2: X = Y + 6

Substitute Y + 6 for X in equation 1:

X - 4 = 2*(Y-4) - 3 becomes Y + 6 - 4 = 2*(Y-4) - 3

That simplifies to Y + 2 = 2Y - 8 - 3

Rewrite it again: Y + 2 = 2Y - 11

Add 11 to both sides: Y + 13 = 2Y

Subtract Y from both sides: 13 = Y

Now plug 13 in for y in Equation 2

X = Y + 6

X = 13 + 6

X = 19

Oct 08, 2009 | Vivendi Math Blaster Algebra (03763) for...

2y - x = 3

x = 3y - 5

Add the two equations side by side,

2y - x + x = 3 + 3y - 5

2y = 3y - 2

y = 2

Plug this in the second equation to get x,

x = 3(2) - 5

x = 1

So the solution is**x = 1**, **y = 2**

or in ordered pair notation**(1, 2)**

x = 3y - 5

Add the two equations side by side,

2y - x + x = 3 + 3y - 5

2y = 3y - 2

y = 2

Plug this in the second equation to get x,

x = 3(2) - 5

x = 1

So the solution is

or in ordered pair notation

Jun 29, 2009 | Computers & Internet

x=7

y=-1

y=-1

Jan 30, 2009 | Bagatrix Algebra Solved! 2005 (105101) for...

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...

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

y=1 x= -3

y=1 x= -3

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

B. 2((2-Y)/3)-Y=3

2(2-Y)-Y=9

4-2Y-Y=9

-3Y=9-4

-3Y=5

-Y=5/3

Y=-1.67

A. 3X+(-1.67)=2

3X-1.67=2

3X=2+1.67

3X=3.67

X=3.67/3

X=1.22

2(2-Y)-Y=9

4-2Y-Y=9

-3Y=9-4

-3Y=5

-Y=5/3

Y=-1.67

A. 3X+(-1.67)=2

3X-1.67=2

3X=2+1.67

3X=3.67

X=3.67/3

X=1.22

Nov 05, 2007 | SoftMath Algebrator - Algebra Homework...

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

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

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6x-5y=47

3y=-7-2x

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