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Verified Answer :

Answer:

Option (2) is correct.

\(2y^2-y-6=0\)

Step-by-step explanation:

Given two equations \(x=y^2-1\) and \(2x-y=4\)

We have to choose an equation from the given options which could be the result of using substitution to solve the system.

Since the given equation is in term of y , so we make substitution of x in term of y.

From \(2x-y=4\),

⇒ \(x=\frac{4+y}{2}\)

Substitute  \(x=\frac{4+y}{2}\)  in other equation \(x=y^2-1\)

We get, \(x=y^2-1 \Rightarrow \frac{4+y}{2}=y^2-1\)

Cross multiply we get,

\(\Rightarrow {4+y}=2y^2-2\)

Solving further, we get

\(\Rightarrow 2y^2-y-6=0\)

Thus, option (2) is correct.

abidemiokin

The equations that could be the result of using substitution to solve the system is 2y^2 - y - 6 = 0

System of equations


Given the system of equations expressed as:

x = y^2 - 1

2x - y = 4

Substitute equation 1 into  to have:

2(y^2 - 1) - y= 4

Expand to have:

2y^2 - 2 - y = 4

2y^2 - y - 6 = 0

Hence the equations that could be the result of using substitution to solve the system is 2y^2 - y - 6 = 0

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