Question
upstudy study bank question image url

Solve the system using elimination. \[ \begin{array}{l}2 x+3 y=-2 \\ 3 x-6 y=18 \\ ([?], \square)\end{array} \]

Ask by Wells Olson. in the United States
Dec 03,2024

Upstudy AI Solution

Tutor-Verified Answer

Answer

The solution is \( (x, y) = (2, -2) \).

Solution

Solve the system of equations \( 2x+3y=-2;3x-6y=18 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}2x+3y=-2\\3x-6y=18\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=-\frac{2+3y}{2}\\3x-6y=18\end{array}\right.\) - step2: Substitute the value of \(x:\) \(3\left(-\frac{2+3y}{2}\right)-6y=18\) - step3: Multiply the terms: \(-\frac{3\left(2+3y\right)}{2}-6y=18\) - step4: Multiply both sides of the equation by LCD: \(\left(-\frac{3\left(2+3y\right)}{2}-6y\right)\times 2=18\times 2\) - step5: Simplify the equation: \(-6-21y=36\) - step6: Move the constant to the right side: \(-21y=36+6\) - step7: Add the numbers: \(-21y=42\) - step8: Change the signs: \(21y=-42\) - step9: Divide both sides: \(\frac{21y}{21}=\frac{-42}{21}\) - step10: Divide the numbers: \(y=-2\) - step11: Substitute the value of \(y:\) \(x=-\frac{2+3\left(-2\right)}{2}\) - step12: Simplify: \(x=2\) - step13: Calculate: \(\left\{ \begin{array}{l}x=2\\y=-2\end{array}\right.\) - step14: Check the solution: \(\left\{ \begin{array}{l}x=2\\y=-2\end{array}\right.\) - step15: Rewrite: \(\left(x,y\right) = \left(2,-2\right)\) The solution to the system of equations using the elimination method is \( (x, y) = (2, -2) \).

Answered by UpStudy AI and reviewed by a Professional Tutor

error msg
Explain
Simplify this solution

Mind Expander

First, let's multiply the first equation by 3 to align the coefficients of \(x\) for easier elimination: \[ 3(2x + 3y) = 3(-2) \implies 6x + 9y = -6 \] Now we have the following system: \[ \begin{array}{l} 6x + 9y = -6 \\ 3x - 6y = 18 \\ \end{array} \] Next, we can multiply the second equation by 2 to match the \(x\) coefficients: \[ 2(3x - 6y) = 2(18) \implies 6x - 12y = 36 \] Now our system looks like this: \[ \begin{array}{l} 6x + 9y = -6 \\ 6x - 12y = 36 \\ \end{array} \] Next, we subtract the first equation from the second equation: \[ (6x - 12y) - (6x + 9y) = 36 - (-6) \] This simplifies to: \[ -12y - 9y = 36 + 6 \] \[ -21y = 42 \] \[ y = -2 \] Now that we have the value for \(y\), we can substitute it back into one of the original equations to find \(x\). Let's use the first equation: \[ 2x + 3(-2) = -2 \] \[ 2x - 6 = -2 \] \[ 2x = 4 \] \[ x = 2 \] Thus, the solution to the system of equations is: \[ (2, -2) \]

Related Questions

Question 4 Consider the following three systems of linear equations. \[ \begin{array}{c} \text { System A } \\ \left\{\begin{array} { c } { \text { System B } } \\ { - 2 x + 3 y = - 7 [ \mathrm { A } 1 ] } \\ { - 7 x + 4 y = 8 } \end{array} \left[\begin{array} { c } { \text { [A2] } } \end{array} \left\{\begin{array} { r } { - 2 x + 3 y = - 7 [ \mathrm { B } 1 ] } \\ { 2 1 x - 1 2 y = - 2 4 [ \mathrm { B } 2 ] } \end{array} \left\{\left\{\begin{array}{c} -2 x+3 y=-7[\mathrm{C} 1] \\ 13 x=-52 \end{array}[\mathrm{C} 2]\right.\right.\right.\right.\right. \end{array} \] Answer the questions below. For each, choose the transformation and then fill in the blank with the correct number. The arrow ( \( \rightarrow \) ) means the expression on the left becomes the expression on the right. (a) How do we transform System A into System B? \( \square \) \( \times \) Equation \( [\mathrm{A} 1] \rightarrow \) Equation \( [ \) B1] \( -3 \times \) Equation [A2] \( \rightarrow \) Equation [B2] \( \square \) \( \times \) Equation \( [\mathrm{A} 1]+ \) Equation \( [\mathrm{A} 2] \rightarrow \) Equation \( [\mathrm{B} 2] \) \( \square \) \( \times \) Equation \( [A 2]+ \) Equation \( [A 1] \rightarrow \) Equation \( [B 1] \) (b) How do we transform System B into System C? \( \times \) Equation [B1] \( \rightarrow \) Equation [C1] \( \square \) \( \times \) Equation [B2] \( \rightarrow \) Equation [C2] \( \square \) \( \times \) Equation \( [\mathrm{B} 1]+ \) Equation \( [\mathrm{B} 2] \rightarrow \) Equation \( [\mathrm{C} 2] \) \( \square \) \( \times \) Equation \( [\mathrm{B} 2]+ \) Equation \( [\mathrm{B} 1] \rightarrow \) Equation \( [\mathrm{C} 1] \)

Latest Algebra Questions

Try Premium now!
Try Premium and ask Thoth AI unlimited math questions now!
Maybe later Go Premium
Study can be a real struggle
Why not UpStudy it?
Select your plan below
Premium

You can enjoy

Start now
  • Step-by-step explanations
  • 24/7 expert live tutors
  • Unlimited number of questions
  • No interruptions
  • Full access to Answer and Solution
  • Full Access to PDF Chat, UpStudy Chat, Browsing Chat
Basic

Totally free but limited

  • Limited Solution
Welcome to UpStudy!
Please sign in to continue the Thoth AI Chat journey
Continue with Email
Or continue with
By clicking “Sign in”, you agree to our Terms of Use & Privacy Policy