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Lyrx [107]
3 years ago
11

Determine which equations below when combined with the equation 3x - 4y equals two or form a system with no solutions. choose al

l that apply.
1. 2y=1.5x-2
2. 2y=1.5-1
3. 3x+4y=2
4. -4y+3x=-2
Mathematics
2 answers:
torisob [31]3 years ago
8 0
The equation given above is, 
           3x - 4y = 2
This can be expressed in slope-intercept form by transposing y to one side of the equation,
      y = (2 - 3x) / -4

Simplifying,
     y = 3x/4 - 1/2

The equations will form a system with no solution only if they have the same slope but different y-intercept.

(1) 2y = 1.5x - 2
      y = 3x/4 - 1
Since this has the same slope as the given equation but different intercept then, this is one of the answers.

(2) 2y = 1.5x - 1
      y = 3x/4 - 0.5

This is exactly the same as the given equation then, the equations will have infinite number of solutions.

(3) 3x + 4y = 2
      4y = (2 - 3x)
       y = -3x/4 - 1/2

The equations do not have the same slope so, this is not one of the answers.

(4) -4y + 3x = -2
     y = (-3x - 2) / -4
     y = 3x/4 - 1/2
This equation is also exactly as that which is given. Hence, they will have infinite number of solutions.
      
ivann1987 [24]3 years ago
6 0

The <em><u>correct answers</u></em> are:

1. 2y=1.5x-2 ; and 4. -4y+3x=-2.

Explanation:

Using the first equation, we have the system

\left \{ {{3x-4y=2} \atop {2y=1.5x-2}} \right.

The second equation can be written in standard form just as the first one.  To do this, we will subtract 1.5x from each side:

2y = 1.5x-2

2y-1.5x = 1.5x-2-1.5x

-1.5x+2y = -2

This makes our system

\left \{ {{3x-4y=2} \atop {-1.5x+2y=-2}} \right.

To solve this, we will make the coefficients of x the same; we do this by multiplying the bottom equation by 2:

2(-1.5x+2y=-2)

-3x+4y=-4

This gives us the system

\left \{ {{3x-4y=2} \atop {-3x+4y=-4}} \right.

We solve this by adding the two equations:

\left \{ {{3x-4y=2} \atop {+(-3x+4y=-4)}} \right. \\\\\\0+0 = -4\\0=-4

There is no solution to this.

For the second system, we will follow the same process, subtract 1.5x from each side of the second equation:

2y=1.5x-1

2y-1.5x = 1.5x-1-1.5x

-1.5x+2y = -1

To make the coefficients of x the same, multiply the second equation by 2:

2(-1.5x+2y=-1)

-3x+4y=-2

We will add the two equations to solve:

\left \{ {{3x-4y=2} \atop {+(-3x+4y=-2)}} \right. \\0+0=0\\0=0

This means that the equations are of the same line and there are infinite solutions.

For the third system, the coefficients are already the same.  We will cancel by subtracting the bottom equation from the top:

\left \{ {{3x-4y=2} \atop {-(3x+4y=2)}} \right. \\\\-4y-4y=2-2\\-8y=0\\\frac{-8y}{-8}=\frac{0}{-8}\\y=0

Since we have a value for y, this has a solution.

For the last system, we will rearrange the second equation with the x term in front:

3x-4y=-2

Now we will subtract this from the first equation:

\left \{ {{3x-4y=2} \atop {-(3x-4y=-2)}} \right. \\-4y--4y=2--2\\0=4

This has no solution.

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