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Sagot :
Sure, let's solve this system of linear equations step-by-step:
[tex]\[ \left\{ \begin{array}{c} 2x - 3y = 10 \\ -5x + 8y = -26 \end{array} \right. \][/tex]
### Step 1: Identify the equations
Equation 1: [tex]\(2x - 3y = 10\)[/tex] \\
Equation 2: [tex]\(-5x + 8y = -26\)[/tex]
### Step 2: Use the method of substitution or elimination
Let's use the elimination method to solve this system of equations. The goal is to eliminate one of the variables by adding or subtracting the equations.
First, we need to make the coefficients of [tex]\(x\)[/tex] or [tex]\(y\)[/tex] the same.
### Step 3: Make the coefficients of [tex]\(x\)[/tex] the same
To do so, we multiply Equation 1 by 5 and Equation 2 by 2:
[tex]\[ 5(2x - 3y) = 5(10) \implies 10x - 15y = 50 \quad \text{(Equation 3)} \][/tex]
[tex]\[ 2(-5x + 8y) = 2(-26) \implies -10x + 16y = -52 \quad \text{(Equation 4)} \][/tex]
### Step 4: Add the equations to eliminate [tex]\(x\)[/tex]
Now, add Equation 3 and Equation 4:
[tex]\[ (10x - 15y) + (-10x + 16y) = 50 + (-52) \][/tex]
This simplifies to:
[tex]\[ 10x - 10x - 15y + 16y = -2 \][/tex]
[tex]\[ y = -2 \][/tex]
### Step 5: Substitute [tex]\(y\)[/tex] back into one of the original equations to find [tex]\(x\)[/tex]
We can use Equation 1 to find [tex]\(x\)[/tex]:
[tex]\[ 2x - 3(-2) = 10 \][/tex]
This simplifies to:
[tex]\[ 2x + 6 = 10 \][/tex]
Subtract 6 from both sides:
[tex]\[ 2x = 4 \][/tex]
Divide by 2:
[tex]\[ x = 2 \][/tex]
### Step 6: Solution
The solution to the system of equations is:
[tex]\[ x = 2, \quad y = -2 \][/tex]
So, the point [tex]\((x, y)\)[/tex] that satisfies both equations is [tex]\((2, -2)\)[/tex].
[tex]\[ \left\{ \begin{array}{c} 2x - 3y = 10 \\ -5x + 8y = -26 \end{array} \right. \][/tex]
### Step 1: Identify the equations
Equation 1: [tex]\(2x - 3y = 10\)[/tex] \\
Equation 2: [tex]\(-5x + 8y = -26\)[/tex]
### Step 2: Use the method of substitution or elimination
Let's use the elimination method to solve this system of equations. The goal is to eliminate one of the variables by adding or subtracting the equations.
First, we need to make the coefficients of [tex]\(x\)[/tex] or [tex]\(y\)[/tex] the same.
### Step 3: Make the coefficients of [tex]\(x\)[/tex] the same
To do so, we multiply Equation 1 by 5 and Equation 2 by 2:
[tex]\[ 5(2x - 3y) = 5(10) \implies 10x - 15y = 50 \quad \text{(Equation 3)} \][/tex]
[tex]\[ 2(-5x + 8y) = 2(-26) \implies -10x + 16y = -52 \quad \text{(Equation 4)} \][/tex]
### Step 4: Add the equations to eliminate [tex]\(x\)[/tex]
Now, add Equation 3 and Equation 4:
[tex]\[ (10x - 15y) + (-10x + 16y) = 50 + (-52) \][/tex]
This simplifies to:
[tex]\[ 10x - 10x - 15y + 16y = -2 \][/tex]
[tex]\[ y = -2 \][/tex]
### Step 5: Substitute [tex]\(y\)[/tex] back into one of the original equations to find [tex]\(x\)[/tex]
We can use Equation 1 to find [tex]\(x\)[/tex]:
[tex]\[ 2x - 3(-2) = 10 \][/tex]
This simplifies to:
[tex]\[ 2x + 6 = 10 \][/tex]
Subtract 6 from both sides:
[tex]\[ 2x = 4 \][/tex]
Divide by 2:
[tex]\[ x = 2 \][/tex]
### Step 6: Solution
The solution to the system of equations is:
[tex]\[ x = 2, \quad y = -2 \][/tex]
So, the point [tex]\((x, y)\)[/tex] that satisfies both equations is [tex]\((2, -2)\)[/tex].
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