For the equation + 2 = 19 and 2 − 3 = − 3 Find the value of D.
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Answer:
Answer:
The value of D for the given simultaneous equations is - 8.
Option ( 1 )
Step-by-step-explanation:
The given simultaneous equations are
\displaystyle{\sf\:y\:+\:2x\:=\:19\:\&\:2x\:-\:3y\:=\:-\:3}y+2x=19&2x−3y=−3
We have to find the value of determinant D.
Now,
\displaystyle{\sf\:y\:+\:2x\:=\:19}y+2x=19
\displaystyle{\implies\sf\:2x\:+\:y\:=\:19\:\:\:-\:-\:(\:1\:)}⟹2x+y=19−−(1)
\displaystyle{\bullet\:\sf\:a_1\:=\:2}∙a
1
=2
\displaystyle{\bullet\:\sf\:b_1\:=\:1}∙b
1
=1
\displaystyle{\bullet\:\sf\:c_1\:=\:19}∙c
1
=19
Now,
\displaystyle{\sf\:2x\:-\:3y\:=\:-\:3\:\:\:-\:-\:(\:2\:)}2x−3y=−3−−(2)
\displaystyle{\bullet\:\sf\:a_2\:=\:2}∙a
2
=2
\displaystyle{\bullet\:\sf\:b_2\:=\:-\:3}∙b
2
=−3
\displaystyle{\bullet\:\sf\:c_2\:=\:-\:3}∙c
2
=−3
Now, we know that,
\begin{gathered}\displaystyle{\pink{\sf\:D\:=\:\left|\begin{array}{cc}\sf\:a_1 & \sf\:b_1\\\sf\:a_2 & \sf\:b_2\:\end{array}\:\right|}}\end{gathered}
D=
∣
∣
∣
∣
∣
a
1
a
2
b
1
b
2
∣
∣
∣
∣
∣
\begin{gathered}\displaystyle{\implies\sf\:D\:=\:\left|\begin{array}{cc}\sf\:2 & \sf\:1\\\sf\:2 & \sf\:-\:3\:\end{array}\:\right|}\end{gathered}
⟹D=
∣
∣
∣
∣
∣
2
2
1
−3
∣
∣
∣
∣
∣
\displaystyle{\implies\sf\:D\:=\:2\:\times\:(\:-\:3\:)\:-\:1\:\times\:2}⟹D=2×(−3)−1×2
\displaystyle{\implies\sf\:D\:=\:-\:6\:-\:2}⟹D=−6−2
\displaystyle{\implies\underline{\boxed{\red{\sf\:D\:=\:-\:8}}}}⟹
D=−8
∴ The value of D for the given simultaneous equations is - 8.