Problem 5

Question

Determine $c$ such that $$ \begin{array}{l} 2 x-3 y=5 \\ 4 x-6 y=c \end{array} $$ has (a) infinitely many solutions and (b) no solutions. (c) Is it possible to choose a number for $c$ so that the system has exactly one solution? Explain your answer.

Step-by-Step Solution

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Answer

(a) Infinitely many solutions when $c = 10$; (b) No solutions when $c \neq 10$; (c) Impossible to have exactly one solution.

1Step 1: Understand the condition for infinitely many solutions

For a system of linear equations to have infinitely many solutions, the two equations must be equivalent. This happens when they are proportional: one equation is a constant multiple of the other. Let's check if the second equation can be a multiple of the first equation.

2Step 2: Determine if the equations are equivalent

The first equation is $2x - 3y = 5$. The second equation is $4x - 6y = c$. Notice that if we multiply the first equation by 2, we obtain $4x - 6y = 10$. For the equations to be equivalent, we need $c = 10$.

3Step 3: Condition for no solutions

For a system to have no solution, the equations must be parallel but not coincident, meaning they have the same slope but different intercepts. Since the second equation is $4x - 6y = c$, it is parallel to the first equation $2x - 3y = 5$. Substituting any value of $c$ that makes $c eq 10$ will result in no solutions.

4Step 4: Condition for exactly one solution

A system of linear equations has exactly one solution if the lines are not parallel, meaning the equations have different slopes. In our system, both equations already have the same slope; therefore, it is not possible to pick a number for $c$ that results in exactly one solution.

Key Concepts

Infinite solutionsNo solutionsSingle solution

Infinite solutions

Infinite solutions occur in a system of equations when the equations represent the same line. This means that each equation in the system is essentially just a scaled version of the other or, in simple terms, they overlap entirely.

For this to happen, any equation in the system should be a constant multiple of another equation. For example, in our given system:

First equation: $2x - 3y = 5$
Second equation: $4x - 6y = c$

If we multiply every term in the first equation by 2, we get $4x - 6y = 10$. Therefore, to have infinitely many solutions, $c$ must equal 10, because this makes the second equation exactly the same as the scaled-up form of the first equation.

In conclusion, if you set $c = 10$, every pair $(x, y)$ that solves the first equation also solves the second, resulting in infinite solutions.

No solutions

A system of linear equations will have no solutions when the equations represent parallel lines. Parallel lines never intersect, meaning there isn’t a point that satisfies both equations.

To determine parallelism, the coefficient ratios of the variables must be the same, but the constants differ, indeed having the same slope but different intercepts.

In our exercise:

First equation: $2x - 3y = 5$
Second equation: $4x - 6y = c$

The coefficients of $x$ and $y$ show these two lines are parallel. However, to ensure they don't overlap, $c$ must be different from 10, the value required to make them equivalent.

By choosing any $c$ that isn't 10, we establish that the second equation can never become the first through scaling, and thus, these lines won’t meet, resulting in no solutions.

Single solution

A single solution in a system of equations arises when the lines intersect at precisely one point. This requires that the lines not be parallel, implying they must have different slopes.

In examining our system of equations:

First equation: $2x - 3y = 5$
Second equation: $4x - 6y = c$

Both equations are set up as parallel lines due to their equivalent coefficients for $x$ and $y$. Since identical slopes make parallel lines, there's no chance for these lines to intersect at a single point, no matter the value of $c$.

In this particular problem setup, it is impossible to adjust $c$ to achieve a scenario where there is exactly one solution.

Problem 4

Problem 5

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Problem 4

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Problem 5

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Problem 6

$$ \text { Show that } 2(A+B)=2 A+2 B \text { . } $$

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