Answered

Use substitution to solve each system of equations.


  1. $\begin{array}{l}

y=5 x+1 \\

4 x+y=10

\end{array}$


  1. $\begin{array}{l}

y=4 x+5 \\

2 x+y=17

\end{array}$


  1. $\begin{array}{l}

y=3 x-34 \\

y=2 x-5

\end{array}$


  1. $\begin{array}{l}

y=3 x-2 \\

y=2 x-5

\end{array}$


  1. $\begin{array}{l}

2 x+y=3 \\

4 x+4 y=8

\end{array}$


  1. $\begin{array}{l}

3 x+4 y=-3 \\

x+2 y=-1

\end{array}$


  1. $\begin{array}{l}

y=-3 x+4 \\

-6 x-2 y=-8

\end{array}$


  1. $\begin{array}{l}

-1=2 x-y \\

8 x-4 y=-4

\end{array}$


  1. $\begin{array}{l}

x=y-1 \\

-x+y=-1

\end{array}$


  1. $\begin{array}{l}

y=-4 x+11 \\

3 x+y=9

\end{array}$


  1. $\begin{array}{l}

y=-3 x+1 \\

2 x+y=1

\end{array}$


  1. $\begin{array}{l}

3 x+y=-5 \\

6 x+2 y=10

\end{array}$


  1. $\begin{array}{l}

5 x-y=5 \\

-x+3 y=13

\end{array}$


  1. $\begin{array}{l}

2 x+y=4 \\

-2 x+y=-4

\end{array}$


  1. $\begin{array}{l}

-5 x+4 y=20 \\

10 x-8 y=-40

\end{array}$

Verified Answer :

  • For each system, isolate one variable and substitute it into the other equation to solve for the first variable.
  • Determine the second variable by substituting the known value back into one of the original equations.
  • Identify systems with no solution or infinitely many solutions by checking for logical consistency (e.g., \(0 = 0\) or a false equality).
  • The final solutions are: 1.\((1, 6)\), 2.\((2, 13)\), 3.\((29, 53)\), 4.\((-3, -11)\), 5.\((1, 1)\), 6.\((-1, 0)\), 7.infinitely many, 8.infinitely many, 9.none, 10.\((2, 3)\), 11.\((0, 1)\), 12.none, 13.\((2, 5)\), 14.\((2, 0)\), 15.infinitely many.

Explanation

  1. Understanding the Methodology

We are tasked with solving 15 systems of linear equations in two variables, \((x, y)\), using the substitution method. This method involves isolating one variable in one equation and substituting it into the other to solve for the remaining variable. A system may yield a single solution \((x, y)\), no solution (when variables cancel to an impossible equality), or infinitely many solutions (when variables cancel to a true equality like \(0 = 0\)).

  1. Solving the Systems
  2. \(\begin{cases} y=5x+1 \ 4x+y=10 \end{cases}\): Substitute \(y = 5x + 1\) into the second: \(4x + (5x + 1) = 10 \Rightarrow 9x = 9 \Rightarrow x = 1\). Then \(y = 5(1) + 1 = 6\). Solution: \((1, 6)\).
  3. \(\begin{cases} y=4x+5 \ 2x+y=17 \end{cases}\): Substitute: \(2x + (4x + 5) = 17 \Rightarrow 6x = 12 \Rightarrow x = 2\). Then \(y = 4(2) + 5 = 13\). Solution: \((2, 13)\).
  4. \(\begin{cases} y=3x-34 \ y=2x-5 \end{cases}\): Substitute: \(3x - 34 = 2x - 5 \Rightarrow x = 29\). Then \(y = 2(29) - 5 = 53\). Solution: \((29, 53)\).
  5. \(\begin{cases} y=3x-2 \ y=2x-5 \end{cases}\): Substitute: \(3x - 2 = 2x - 5 \Rightarrow x = -3\). Then \(y = 2(-3) - 5 = -11\). Solution: \((-3, -11)\).
  6. \(\begin{cases} 2x+y=3 \ 4x+4y=8 \end{cases}\): From first, \(y = 3 - 2x\). Substitute into second: \(4x + 4(3 - 2x) = 8 \Rightarrow 4x + 12 - 8x = 8 \Rightarrow -4x = -4 \Rightarrow x = 1\). Then \(y = 3 - 2(1) = 1\). Solution: \((1, 1)\).
  7. \(\begin{cases} 3x+4y=-3 \ x+2y=-1 \end{cases}\): From second, \(x = -1 - 2y\). Substitute: \(3(-1 - 2y) + 4y = -3 \Rightarrow -3 - 6y + 4y = -3 \Rightarrow -2y = 0 \Rightarrow y = 0\). Then \(x = -1 - 2(0) = -1\). Solution: \((-1, 0)\).
  8. \(\begin{cases} y=-3x+4 \ -6x-2y=-8 \end{cases}\): Substitute: \(-6x - 2(-3x + 4) = -8 \Rightarrow -6x + 6x - 8 = -8 \Rightarrow -8 = -8\). Infinitely many solutions.
  9. \(\begin{cases} -1=2x-y \ 8x-4y=-4 \end{cases}\): From first, \(y = 2x + 1\). Substitute: \(8x - 4(2x + 1) = -4 \Rightarrow 8x - 8x - 4 = -4 \Rightarrow -4 = -4\). Infinitely many solutions.
  10. \(\begin{cases} x=y-1 \ -x+y=-1 \end{cases}\): Substitute: \(-(y - 1) + y = -1 \Rightarrow -y + 1 + y = -1 \Rightarrow 1 = -1\). No solution.
  11. \(\begin{cases} y=-4x+11 \ 3x+y=9 \end{cases}\): Substitute: \(3x + (-4x + 11) = 9 \Rightarrow -x = -2 \Rightarrow x = 2\). Then \(y = -4(2) + 11 = 3\). Solution: \((2, 3)\).
  12. \(\begin{cases} y=-3x+1 \ 2x+y=1 \end{cases}\): Substitute: \(2x + (-3x + 1) = 1 \Rightarrow -x = 0 \Rightarrow x = 0\). Then \(y = 1\). Solution: \((0, 1)\).
  13. \(\begin{cases} 3x+y=-5 \ 6x+2y=10 \end{cases}\): From first, \(y = -5 - 3x\). Substitute: \(6x + 2(-5 - 3x) = 10 \Rightarrow 6x - 10 - 6x = 10 \Rightarrow -10 = 10\). No solution.
  14. \(\begin{cases} 5x-y=5 \ -x+3y=13 \end{cases}\): From first, \(y = 5x - 5\). Substitute: \(-x + 3(5x - 5) = 13 \Rightarrow -x + 15x - 15 = 13 \Rightarrow 14x = 28 \Rightarrow x = 2\). Then \(y = 5(2) - 5 = 5\). Solution: \((2, 5)\).
  15. \(\begin{cases} 2x+y=4 \ -2x+y=-4 \end{cases}\): From first, \(y = 4 - 2x\). Substitute: \(-2x + (4 - 2x) = -4 \Rightarrow -4x = -8 \Rightarrow x = 2\). Then \(y = 4 - 2(2) = 0\). Solution: \((2, 0)\).
  16. \(\begin{cases} -5x+4y=20 \ 10x-8y=-40 \end{cases}\): From first, \(4y = 5x + 20 \Rightarrow y = 1.25x + 5\). Substitute: \(10x - 8(1.25x + 5) = -40 \Rightarrow 10x - 10x - 40 = -40 \Rightarrow -40 = -40\). Infinitely many solutions.
  1. Final Results

The solutions for the systems are:

  1. \((1, 6)\), 2. \((2, 13)\), 3. \((29, 53)\), 4. \((-3, -11)\), 5. \((1, 1)\), 6. \((-1, 0)\), 7. Infinitely many solutions, 8. Infinitely many solutions, 9. No solution, 10. \((2, 3)\), 11. \((0, 1)\), 12. No solution, 13. \((2, 5)\), 14. \((2, 0)\), 15. Infinitely many solutions.

Examples

In urban planning, solving systems of linear equations helps in balancing traffic flow. If two roads intersect at specific points represented by the variables x and y, finding the solution to a system of equations determines exactly where these roads meet. For example, if road A is described by \(y = 5x + 1\) and road B by \(4x + y = 10\), the intersection point \((1, 6)\) tells planners where traffic signals or roundabouts should be installed to manage the flow effectively. This fundamental algebraic method is essential for infrastructure design and resource allocation.

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