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# hi how can i solve it ?-2x+3y-12=0i know that x*y=90

i don't know how to solve it ?

### 4 Answers by Expert Tutors

Roman C. | Masters of Education Graduate with Mathematics ExpertiseMasters of Education Graduate with Mathe...
4.9 4.9 (350 lesson ratings) (350)
1

Very quick way.

Multiply the first equation by x to get

-2x2 + 3xy - 12x = 0

Now use the second equation xy = 90 to substitute and get

-2x2 + 3*90 - 12x = 0

-2x2 - 12x + 270 = 0

-2(x + 15)(x - 9) = 0

x = -15 or 9

Using xy = 90, the two solutions (x,y) are (-15,-6) and (9,10)

Roman, this was clever, and it cleaned up the problem very quickly.
Nice job.

BRUCE S. | Learn & Master Physics & Math with Bruce SLearn & Master Physics & Math with Bruce...
4.9 4.9 (36 lesson ratings) (36)
0

Omid,

Problem information is:

-2x+3y-12=0    eqn 1

x*y=90            eqn 2

isolate x from eqn 2:

x=90/y             eqn 3

Substitute x result from eqn 3 into eqn 1

-2(90/y) +3y -12 =0

Multiply the above by y and collect like terms:

-180 +3y^2 -12y =0

3y^2 -12y -180 =0

Divide the above by 3:

y^2 -4y -60 =0

factor this simplified result to:

(y+6)*(y-10)

Note: Check this using FOIL.

Therefore:  y=-6 or +10    eqn 5

Substitute eqn 5 results back into eqn 1 to find the x values:

x*y = 90

y=-6 =>   x*(-6) = 90   thus x = -15

y=10 =>  x*10 = 90    thus x = 9

Final results:

x= -15 or 9

y= -6 or 10

BruceS

Sheila M. | Friendly College Math Professor - understands students' math anxietyFriendly College Math Professor - unders...
4.9 4.9 (63 lesson ratings) (63)
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Hi, Omid.

It looks like you're learning about solving systems of equations.  The easiest way to approach this particular problem is to use substitution.  In order to do that, you need to get one of the variables by itself.  I think it is easier to work with the first equation so that we can avoid mistakes made by using fractions.

-2x + 3y - 12 = 0

Let's get y alone.

-2x    +   3y   - 12   =   0
+2x                           +2x
3y  -  12   =   2x

3y   - 12   = 2x
+ 12            +12
3y           =  2x  + 12

Now divide each term on both sides of the equation by 3

3y = 2x +  12
3      3      3

y = 2/3 x   + 4

Now that we have y alone in the first equation, we can substitute its expression in the other equation.  Here we go:

x * y = 90

x * (2/3 x  + 4)  = 90

Distribute the x...

2/3 x^2  +  4x  = 90

Let's multiply all the terms (on both sides) by 3 to clear the fractions:

2 x^2  + 12x = 270

Do you see that we now have a quadratic equation to solve?
Let's get all the terms on one side first:

2 x^2    +   12x         =   270
-270         -270
2 x^2   + 12x  - 270   =    0

It's easiest to factor if the coefficient of the x^2 term is a 1.  Since all four terms are divisible by 2, let's divide both sides by 2 (all the terms):

2 x^2   + 12x   -   270  =   0
2           2           2         2

x^2  + 6x   - 135  = 0

Do you know how to factor this?
When you get your two answers, be sure to test them in the original two equations to ensure they are solutions.

Kevin S. |
5.0 5.0 (4 lesson ratings) (4)
0

If x*y = 90, you can solve it for x (or y) and substitute into the other equations:

y = 90/x

-2x + 3 (90/x) - 12 = 0

Now multiply everything by x to move the x from the denominator:

-2x2 + 270 - 12x = 0

Reorder terms -2x2 - 12x + 270 = 0

Now factor:  -2x2 - 12x + 270 = 0

-2 (x2 + 6x - 135) = 0

Now we can divide out the -2 because it will never equal zero, and factor inside the parentheses:

(x2 + 6x - 135) = 0

Now, what factors of 135, when one is subtracted from the other, equal 6? Let's try 15 and 9

(x + 15      )(x -9) = 0

So x + 15 = 0 or x = -15
and x -9 = 0 or x = 9

Since x * y = 90, then (9)y = 90  ===> y = 10
and (-15)y = 90 ===> y = -6

so your solutions are (9,10) and (-15, -6)