CALCU-Question \#1: Create a system of linear equa

Solution by Steps

step 1

To create a system of linear equations with no solutions, we need two lines that are parallel but not coincident. For example, consider the equations:

y = 2x + 3

y = 2x + 5

These lines have the same slope but different y-intercepts, so they will never intersect

step 2

To create a system of linear equations with one solution, we need two lines that intersect at exactly one point. For example, consider the equations:

y = 2x + 3

y = -x + 1

These lines have different slopes, so they will intersect at exactly one point

step 3

To create a system of linear equations with infinitely many solutions, we need two lines that are coincident. For example, consider the equations:

y = 2x + 3

2y = 4x + 6

These lines are actually the same line, so they will intersect at every point on the line

step 4

To verify the solutions algebraically, we solve each system of equations. For the system with no solutions:

y = 2x + 3

y = 2x + 5

Subtracting the first equation from the second:

0 = 2

This is a contradiction, so there are no solutions

step 5

For the system with one solution:

y = 2x + 3

y = -x + 1

Setting the equations equal to each other:

2x + 3 = -x + 1

3x = -2

x = -\frac{2}{3}

Substituting back into the first equation:

y = 2(-\frac{2}{3}) + 3

y = -\frac{4}{3} + 3

y = \frac{5}{3}

So the solution is

\left( -\frac{2}{3}, \frac{5}{3} \right)

step 6

For the system with infinitely many solutions:

y = 2x + 3

2y = 4x + 6

Dividing the second equation by 2:

y = 2x + 3

This is the same as the first equation, so there are infinitely many solutions

step 7

To verify the solutions graphically, we plot each system of equations on a graph. For the system with no solutions, the lines will be parallel and never intersect. For the system with one solution, the lines will intersect at exactly one point. For the system with infinitely many solutions, the lines will be coincident and overlap completely

Answer

No solutions:

y = 2x + 3

and

y = 2x + 5

One solution:

y = 2x + 3

and

y = -x + 1

Infinitely many solutions:

y = 2x + 3

and

2y = 4x + 6

Key Concept

System of Linear Equations

Explanation

A system of linear equations can have no solutions (parallel lines), one solution (intersecting lines), or infinitely many solutions (coincident lines). The number of solutions can be determined by comparing the slopes and y-intercepts of the lines.

Generated Graph

Solution by Steps

step 1

To factor the expression

64a^6 - 25

, we recognize it as a difference of squares:

64a^6 - 25 = (8a^3)^2 - 5^2

step 2

Using the difference of squares formula,

a^2 - b^2 = (a - b)(a + b)

, we get:

(8a^3 - 5)(8a^3 + 5)

step 3

Therefore, the fully factored form of

64a^6 - 25

is

(8a^3 - 5)(8a^3 + 5)

step 4

To factor the expression

4x^2y - 44xy + 72y

, we first factor out the common term

4y

:

4y(x^2 - 11x + 18)

step 5

Next, we factor the quadratic

x^2 - 11x + 18

. We look for two numbers that multiply to

18

and add to

-11

, which are

-2

and

-9

step 6

Thus,

x^2 - 11x + 18 = (x - 2)(x - 9)

step 7

Therefore, the fully factored form of

4x^2y - 44xy + 72y

is

4y(x - 2)(x - 9)

step 8

To factor the expression

6n^2 - 11ny - 10y^2

, we look for two numbers that multiply to

6 \cdot (-10) = -60

and add to

-11

. These numbers are

-15

and

4

step 9

We rewrite the middle term using these numbers:

6n^2 - 15ny + 4ny - 10y^2

step 10

We factor by grouping:

3n(2n - 5y) + 2y(2n - 5y)

step 11

We factor out the common binomial factor:

(3n + 2y)(2n - 5y)

step 12

Therefore, the fully factored form of

6n^2 - 11ny - 10y^2

is

(3n + 2y)(2n - 5y)

[3] Answer

A

Key Concept

Factoring Polynomials

Explanation

Factoring polynomials involves recognizing patterns such as the difference of squares, common factors, and factoring by grouping.

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