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Sagot :
Let's solve the given system of linear equations step-by-step:
[tex]\[ \begin{aligned} 1) \quad &-5x + 2y = -10 \quad &\text{(Equation 1)} \\ 2) \quad &3x - 6y = -18 \quad &\text{(Equation 2)} \end{aligned} \][/tex]
Step 1: Identify a variable to eliminate
To eliminate a variable, we usually choose the one that will result in simpler arithmetic. Let's eliminate [tex]\(y\)[/tex] first because the coefficients of [tex]\(y\)[/tex] in both equations are easier to manipulate.
Step 2: Make the coefficients of [tex]\(y\)[/tex] equal
To eliminate [tex]\(y\)[/tex], we need to make the coefficients of [tex]\(y\)[/tex] in both equations equal in magnitude. The coefficients of [tex]\(y\)[/tex] are 2 in Equation 1 and -6 in Equation 2. To make the coefficients of [tex]\(y\)[/tex] equal, we can multiply Equation 1 by 3:
[tex]\[ 3(-5x + 2y) = 3(-10) \][/tex]
This gives us:
[tex]\[ -15x + 6y = -30 \quad \text{(Modified Equation 1)} \][/tex]
Now we have:
[tex]\[ \begin{aligned} -15x + 6y &= -30 \quad &\text{(Modified Equation 1)} \\ 3x - 6y &= -18 \quad &\text{(Equation 2)} \end{aligned} \][/tex]
Step 3: Add the equations to eliminate [tex]\(y\)[/tex]
Next, we add the Modified Equation 1 and Equation 2 together to eliminate [tex]\(y\)[/tex]:
[tex]\[ (-15x + 6y) + (3x - 6y) = -30 + (-18) \][/tex]
This simplifies to:
[tex]\[ -15x + 3x + 6y - 6y = -48 \][/tex]
[tex]\[ -12x = -48 \][/tex]
Step 4: Solve for [tex]\(x\)[/tex]
Now, we solve for [tex]\(x\)[/tex]:
[tex]\[ x = \frac{-48}{-12} = 4 \][/tex]
Step 5: Substitute [tex]\(x = 4\)[/tex] back into one of the original equations to find [tex]\(y\)[/tex]
We can substitute [tex]\(x = 4\)[/tex] into either Equation 1 or Equation 2. Let's use Equation 1:
[tex]\[ -5(4) + 2y = -10 \][/tex]
This simplifies to:
[tex]\[ -20 + 2y = -10 \][/tex]
Adding 20 to both sides:
[tex]\[ 2y = -10 + 20 \][/tex]
[tex]\[ 2y = 10 \][/tex]
Solving for [tex]\(y\)[/tex]:
[tex]\[ y = \frac{10}{2} = 5 \][/tex]
Final Answer:
The solution to the system of equations is:
[tex]\[ x = 4, \quad y = 5 \][/tex]
We eliminated the variable [tex]\(y\)[/tex] first by making the coefficients of [tex]\(y\)[/tex] equal in both equations, and then adding them to eliminate [tex]\(y\)[/tex].
[tex]\[ \begin{aligned} 1) \quad &-5x + 2y = -10 \quad &\text{(Equation 1)} \\ 2) \quad &3x - 6y = -18 \quad &\text{(Equation 2)} \end{aligned} \][/tex]
Step 1: Identify a variable to eliminate
To eliminate a variable, we usually choose the one that will result in simpler arithmetic. Let's eliminate [tex]\(y\)[/tex] first because the coefficients of [tex]\(y\)[/tex] in both equations are easier to manipulate.
Step 2: Make the coefficients of [tex]\(y\)[/tex] equal
To eliminate [tex]\(y\)[/tex], we need to make the coefficients of [tex]\(y\)[/tex] in both equations equal in magnitude. The coefficients of [tex]\(y\)[/tex] are 2 in Equation 1 and -6 in Equation 2. To make the coefficients of [tex]\(y\)[/tex] equal, we can multiply Equation 1 by 3:
[tex]\[ 3(-5x + 2y) = 3(-10) \][/tex]
This gives us:
[tex]\[ -15x + 6y = -30 \quad \text{(Modified Equation 1)} \][/tex]
Now we have:
[tex]\[ \begin{aligned} -15x + 6y &= -30 \quad &\text{(Modified Equation 1)} \\ 3x - 6y &= -18 \quad &\text{(Equation 2)} \end{aligned} \][/tex]
Step 3: Add the equations to eliminate [tex]\(y\)[/tex]
Next, we add the Modified Equation 1 and Equation 2 together to eliminate [tex]\(y\)[/tex]:
[tex]\[ (-15x + 6y) + (3x - 6y) = -30 + (-18) \][/tex]
This simplifies to:
[tex]\[ -15x + 3x + 6y - 6y = -48 \][/tex]
[tex]\[ -12x = -48 \][/tex]
Step 4: Solve for [tex]\(x\)[/tex]
Now, we solve for [tex]\(x\)[/tex]:
[tex]\[ x = \frac{-48}{-12} = 4 \][/tex]
Step 5: Substitute [tex]\(x = 4\)[/tex] back into one of the original equations to find [tex]\(y\)[/tex]
We can substitute [tex]\(x = 4\)[/tex] into either Equation 1 or Equation 2. Let's use Equation 1:
[tex]\[ -5(4) + 2y = -10 \][/tex]
This simplifies to:
[tex]\[ -20 + 2y = -10 \][/tex]
Adding 20 to both sides:
[tex]\[ 2y = -10 + 20 \][/tex]
[tex]\[ 2y = 10 \][/tex]
Solving for [tex]\(y\)[/tex]:
[tex]\[ y = \frac{10}{2} = 5 \][/tex]
Final Answer:
The solution to the system of equations is:
[tex]\[ x = 4, \quad y = 5 \][/tex]
We eliminated the variable [tex]\(y\)[/tex] first by making the coefficients of [tex]\(y\)[/tex] equal in both equations, and then adding them to eliminate [tex]\(y\)[/tex].
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