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2. What is the \( x \) value in the solution for this system of equations? \[ \left\{\begin{array}{c}2 x-y=-5 \\ -3 x-y=-1\end{array}\right. \]

Ask by Beck Reed. in the United States
Jan 27,2025

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Tutor-Verified Answer

Answer

\( x = -\dfrac{4}{5} \)

Solution

To solve the system of equations: \[ \begin{cases} 2x - y = -5 \\ -3x - y = -1 \end{cases} \] **Step 1: Subtract the second equation from the first to eliminate \( y \).** \[ (2x - y) - (-3x - y) = -5 - (-1) \] \[ 2x - y + 3x + y = -5 + 1 \] \[ 5x = -4 \] **Step 2: Solve for \( x \).** \[ x = \frac{-4}{5} \] **Final Answer:** \( x = -\dfrac{4}{5} \)

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To solve the system of equations, you can use either substitution or elimination. Let's use elimination in this case. First, we'll subtract the first equation from the second equation: \[ (-3x - y) - (2x - y) = -1 - (-5) \] This simplifies to: \[ -5x = 4 \] Now, divide both sides by -5: \[ x = -\frac{4}{5} \] So, the value of \( x \) in the solution for this system of equations is \( -\frac{4}{5} \). For the value of \( y \), you can substitute \( x \) back into one of the original equations. Using \( 2x - y = -5 \): \[ 2(-\frac{4}{5}) - y = -5 \] This simplifies to: \[ -\frac{8}{5} - y = -5 \] To solve for \( y \), first convert -5 to a fraction with a denominator of 5: \[ -\frac{8}{5} - y = -\frac{25}{5} \] Now, add \( \frac{8}{5} \) to both sides: \[ -y = -\frac{25}{5} + \frac{8}{5} \] \[ -y = -\frac{17}{5} \] Thus: \[ y = \frac{17}{5} \] So, the solution to the system of equations is \( x = -\frac{4}{5} \) and \( y = \frac{17}{5} \).

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7. Efectueaza sis scrie rezultatul sub formă de putere: \( \begin{array}{lll}\text { a) } \frac{18}{5} \cdot\left(\frac{18}{5}\right)^{2}= & \text { b) }\left(\frac{6}{5}\right)^{2} \cdot\left(\frac{6}{5}\right)^{3} \cdot \frac{6}{5}= & \text { c) }\left(\frac{19}{5}\right)^{5} \cdot\left(\frac{19}{5}\right)^{16}= \\ \begin{array}{lll}\text { d) } \frac{3}{2} \cdot\left(\frac{3}{2}\right)^{3} \cdot\left(\frac{3}{2}\right)^{0} \cdot\left(\frac{3}{2}\right)^{4}= & \text { e) }\left[\left(\frac{28}{5}\right)^{2}\right]^{3}= & \text { f) }\left[\left(\frac{5}{6}\right)^{6}\right]^{7}= \\ \text { g) }\left[\left(\frac{24}{5}\right)^{2} \cdot\left(\frac{24}{5}\right)^{3}\right]^{8}= & \text { h) }\left[\frac{5}{7} \cdot\left(\frac{5}{7}\right)^{0} \cdot\left(\frac{5}{7}\right)^{4}\right]^{5}= & \text { i) }\left(\frac{29}{10}\right)^{10}:\left(\frac{29}{10}\right)^{7}=\end{array} \\ \left.\left.\begin{array}{lll}\text { j) }\left(\frac{1}{3}\right)^{17}: \frac{1}{3}= & \left.\text { k) }\left(\frac{3}{7}\right)^{11} \cdot\left(\frac{9}{49}\right)^{3}:\left(\frac{3}{7}\right)^{15}=1\right)\end{array}\right]\left(1 \frac{1}{2}\right)^{2}\right]^{8}:\left(\frac{3}{2}\right)^{13}= \\ \text { m) }\left(\frac{9}{10}\right)^{7} \cdot\left(\frac{1}{5}\right)^{7}= & \text { n) }\left(\frac{5}{2}\right)^{10} \cdot\left(\frac{8}{5}\right)^{10}: 2^{10}= & \text { o) } 9^{3} \cdot\left(\frac{7}{10}\right)^{3}:\left(\frac{63}{10}\right)^{3}= \\ \text { p) }\left[\left(\frac{1}{5}\right)^{7}\right]^{2} \cdot 6^{14}:\left(\frac{6}{5}\right)^{14}= & \text { q) }\left(\frac{5}{2}\right)^{7}:\left(\frac{5}{2}\right)^{5}= & \end{array} \)
Algebra Romania Jan 30, 2025
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