System A System B{[6x−5y=1]},{[4x−3y=0]}{[−2x+2y=−1]},{[−2x+2y=−1]}How can we get System B from System A ?Choose 1 answer:A) Replace one equation with the sum/difference of both equationsB) Replace only the left-hand side of one equation with the sum/difference of the left-hand sides of both equationsC) Replace one equation with a multiple of itselfD) Replace one equation with a multiple of the other equation
Q. System A System B{[6x−5y=1]},{[4x−3y=0]}{[−2x+2y=−1]},{[−2x+2y=−1]}How can we get System B from System A ?Choose 1 answer:A) Replace one equation with the sum/difference of both equationsB) Replace only the left-hand side of one equation with the sum/difference of the left-hand sides of both equationsC) Replace one equation with a multiple of itselfD) Replace one equation with a multiple of the other equation
Analyze System A and System B: Analyze System A and System B to identify differences and similarities.System A: 1. 6x−5y=12. −2x+2y=−1System B:1. 4x−3y=02. −2x+2y=−1
Focus on Transformation: Notice that the second equation in both systems is identical. Focus on transforming the first equation of System A to match the first equation of System B.
Compare Coefficients: Compare the coefficients of x and y in the first equations of both systems:System A: 6x−5y=1System B: 4x−3y=0We need to find a transformation that changes 6x−5y=1 to 4x−3y=0.
Find Scalar Multiple: Attempt to find a scalar multiple that can be applied to the first equation of System A to achieve the first equation of System B.Let's try multiplying the first equation of System A by a factor k:k(6x−5y)=k(1)We need k(6x−5y) to equal 4x−3y.
Solve for k: Solve for k:k×6x=4x and k×−5y=−3yk=64=32 and k=−5−3=53These values of k are not consistent, indicating a mistake in the approach or calculation.
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