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Find the value of a so that the system has no solutions.

 

y = 11x + 14

2y - 4 = ax + 7 - 3y + x

 May 3, 2022
 #1
avatar+9310 
+1

Collecting like terms,

\(\begin{cases}11x - y = -14\\(a + 1)x - 5y = -11\end{cases}\)

 

Use method of elimination, you should be able to get \(x = \dfrac{59}{a - 54}, y =\dfrac{14a - 107}{a - 54}\).

It does not make sense if the denominator is 0, because you can't divide by 0. Then the system has no solutions when the denominator is 0. Can you continue from here?

 May 3, 2022
 #2
avatar+1365 
+1

We can rewrite the second equation as: \(5y=ax+11+x\)

 

We then factor this into: \(5y=x(a+1)+11\)

 

We now multiply the first equation by 5 to get: \(5y = 55x+ 14 \)

 

Recall that for a system of equations to have no solutions, the two lines must have the same slope. 

 

The slope is the co-efficient of x, so we have the equation: \(55 = 1+a\), where we can solve for a. 

 

Can you take it from here?

 May 4, 2022

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