Answer :
Answer:
Option (2) is correct.
[tex]2y^2-y-6=0[/tex]
Step-by-step explanation:
Given two equations [tex]x=y^2-1[/tex] and [tex]2x-y=4[/tex]
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 [tex]2x-y=4[/tex],
⇒ [tex]x=\frac{4+y}{2}[/tex]
Substitute [tex]x=\frac{4+y}{2}[/tex] in other equation [tex]x=y^2-1[/tex]
We get, [tex]x=y^2-1 \Rightarrow \frac{4+y}{2}=y^2-1[/tex]
Cross multiply we get,
[tex]\Rightarrow {4+y}=2y^2-2[/tex]
Solving further, we get
[tex]\Rightarrow 2y^2-y-6=0[/tex]
Thus, option (2) is correct.
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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