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{:[-x+(1)/(2)y=-3],[0=cx-2y+10]:}
In the system of equations,
c is a constant. For what value of
c does the system of linear equations have no solutions?

$\begin{array}{l} -x+\frac{1}{2} y=-3 \\ 0=c x-2 y+10 \end{array}$ $\newline$ In the system of equations, $c$ is a constant. For what value of $c$ does the system of linear equations have no solutions?

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Solve a system of equations using any method: word problems

Full solution

Q.

\begin{array}{l} -x+\frac{1}{2} y=-3 \\ 0=c x-2 y+10 \end{array}

\newline

In the system of equations,

c

is a constant. For what value of

c

does the system of linear equations have no solutions?

Given System of Equations: We are given a system of two linear equations: $\newline$ $1$ ) $-x + \frac{1}{2}y = -3$ $\newline$ $2$ ) $0 = cx - 2y + 10$ $\newline$ To determine the value of $c$ for which there are no solutions, we need to look at the coefficients of $x$ and $y$ in both equations. If the ratios of the coefficients of $x$ and $y$ are the same, but the constant terms are different, the lines are parallel and there are no solutions. $\newline$ Let's write the first equation in standard form $(Ax + By = C)$ : $\newline$ $-x + \frac{1}{2}y = -3$ $\newline$ Multiply by $2$ to get rid of the fraction: $\newline$ $0 = cx - 2y + 10$ $0$ $\newline$ Now the coefficients for $x$ and $y$ are $0 = cx - 2y + 10$ $3$ and $0 = cx - 2y + 10$ $4$ , respectively.
Determining Parallel Lines: Next, let's write the second equation in standard form as well: $\newline$ $0 = cx - 2y + 10$ $\newline$ Rearrange the terms to get: $\newline$ $cx - 2y = -10$ $\newline$ Now the coefficients for $x$ and $y$ are $c$ and $-2$ , respectively.
Writing Equations in Standard Form: For the system to have no solutions, the lines represented by the equations must be parallel. This means the ratios of the coefficients of $x$ and $y$ must be the same for both equations. So we set up the following proportion using the coefficients from the standard forms of both equations: $\newline$ $-\frac{2}{c} = \frac{1}{(-2)}$ $\newline$ Cross-multiply to solve for $c$ : $\newline$ $-2 \times (-2) = 1 \times c$ $\newline$ $4 = c$
Setting up Proportion: We have found that when $c = 4$ , the coefficients of $x$ and $y$ in both equations have the same ratio, which means the lines are parallel. Therefore, for $c = 4$ , the system of equations has no solutions.

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