Answer :
Sure, let's solve the given system of linear equations step by step using the method of elimination or substitution.
We are given the system:
[tex]\[ \begin{cases} 3x + 2y = -1 \\ 2x - 5y = 31 \end{cases} \][/tex]
### Step 1: Eliminate one of the variables
We can eliminate [tex]\( y \)[/tex] by finding common coefficients for [tex]\( y \)[/tex] in both equations, and then subtracting or adding the equations.
Let's multiply the first equation by 5 and the second equation by 2 so that the coefficients of [tex]\( y \)[/tex] in both equations are opposites:
[tex]\[ \begin{cases} 5(3x + 2y) = 5(-1) \\ 2(2x - 5y) = 2(31) \end{cases} \][/tex]
This simplifies to:
[tex]\[ \begin{cases} 15x + 10y = -5 \\ 4x - 10y = 62 \end{cases} \][/tex]
### Step 2: Add the equations to eliminate [tex]\( y \)[/tex]
Now we add the two new equations together:
[tex]\[ (15x + 10y) + (4x - 10y) = -5 + 62 \][/tex]
Simplifying this:
[tex]\[ 15x + 4x + 10y - 10y = 57 \][/tex]
This reduces to:
[tex]\[ 19x = 57 \][/tex]
### Step 3: Solve for [tex]\( x \)[/tex]
Divide both sides by 19 to solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{57}{19} \][/tex]
[tex]\[ x = 3 \][/tex]
### Step 4: Substitute [tex]\( x \)[/tex] back into one of the original equations
Now that we have [tex]\( x = 3 \)[/tex], substitute it back into the first original equation to solve for [tex]\( y \)[/tex]:
[tex]\[ 3(3) + 2y = -1 \][/tex]
[tex]\[ 9 + 2y = -1 \][/tex]
Subtract 9 from both sides:
[tex]\[ 2y = -1 - 9 \][/tex]
[tex]\[ 2y = -10 \][/tex]
Divide both sides by 2:
[tex]\[ y = \frac{-10}{2} \][/tex]
[tex]\[ y = -5 \][/tex]
### Solution
Thus, the solution to the system of equations is:
[tex]\[ x = 3,\, y = -5 \][/tex]
So, the solution in coordinate form is:
[tex]\[ (3, -5) \][/tex]
We are given the system:
[tex]\[ \begin{cases} 3x + 2y = -1 \\ 2x - 5y = 31 \end{cases} \][/tex]
### Step 1: Eliminate one of the variables
We can eliminate [tex]\( y \)[/tex] by finding common coefficients for [tex]\( y \)[/tex] in both equations, and then subtracting or adding the equations.
Let's multiply the first equation by 5 and the second equation by 2 so that the coefficients of [tex]\( y \)[/tex] in both equations are opposites:
[tex]\[ \begin{cases} 5(3x + 2y) = 5(-1) \\ 2(2x - 5y) = 2(31) \end{cases} \][/tex]
This simplifies to:
[tex]\[ \begin{cases} 15x + 10y = -5 \\ 4x - 10y = 62 \end{cases} \][/tex]
### Step 2: Add the equations to eliminate [tex]\( y \)[/tex]
Now we add the two new equations together:
[tex]\[ (15x + 10y) + (4x - 10y) = -5 + 62 \][/tex]
Simplifying this:
[tex]\[ 15x + 4x + 10y - 10y = 57 \][/tex]
This reduces to:
[tex]\[ 19x = 57 \][/tex]
### Step 3: Solve for [tex]\( x \)[/tex]
Divide both sides by 19 to solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{57}{19} \][/tex]
[tex]\[ x = 3 \][/tex]
### Step 4: Substitute [tex]\( x \)[/tex] back into one of the original equations
Now that we have [tex]\( x = 3 \)[/tex], substitute it back into the first original equation to solve for [tex]\( y \)[/tex]:
[tex]\[ 3(3) + 2y = -1 \][/tex]
[tex]\[ 9 + 2y = -1 \][/tex]
Subtract 9 from both sides:
[tex]\[ 2y = -1 - 9 \][/tex]
[tex]\[ 2y = -10 \][/tex]
Divide both sides by 2:
[tex]\[ y = \frac{-10}{2} \][/tex]
[tex]\[ y = -5 \][/tex]
### Solution
Thus, the solution to the system of equations is:
[tex]\[ x = 3,\, y = -5 \][/tex]
So, the solution in coordinate form is:
[tex]\[ (3, -5) \][/tex]
