Math  /  Algebra

Question```latex Three systems of linear equations.
System B{5x+2y=4[B1]13x=26[B2]System C{5x+2y=4[C1]x=2[C2] \begin{array}{c} \text{System B} \\ \left\{ \begin{array}{c} -5x + 2y = 4 \quad [\mathrm{B}1] \\ -13x = 26 \quad [\mathrm{B}2] \end{array} \right. \end{array} \quad \begin{array}{c} \text{System C} \\ \left\{ \begin{array}{c} -5x + 2y = 4 \quad [\mathrm{C}1] \\ x = -2 \quad [\mathrm{C}2] \end{array} \right. \end{array}
Below, transformation and then fill in the
How do we transform System B into System C?
×\times Equation [B1][B 1] \rightarrow Equation [C1]
Equation [B2][B 2] \rightarrow Equation [C2]
Equation [B1]+[B 1]+ Equation [B2][B 2] \rightarrow Equation [C2][C 2]
Equation [B2]+[B 2]+ Equation [B1][B 1] \rightarrow Equation [C1]

Studdy Solution
Determine the transformation from System B to System C: - Equation [C1][C1] is identical to Equation [B1][B1], so no transformation is needed for this equation. - To transform Equation [B2][B2] into Equation [C2][C2], solve 13x=26-13x = 26 for xx: $ -13x = 26 \implies x = \frac{26}{-13} = -2 \] This matches Equation \([C2]\).
The transformation from System B to System C involves recognizing that Equation [B1][B1] is already in the form of Equation [C1][C1] and solving Equation [B2][B2] to find x=2x = -2, which is Equation [C2][C2].
The correct transformation is: - Equation [B1][B1] becomes Equation [C1][C1]. - Solving Equation [B2][B2] gives Equation [C2][C2].

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