Answer :
Let's solve the system of equations using the substitution method. The system of equations is:
[tex]\[ \begin{aligned} 5x + 4y &= 7 \quad \text{(Equation 1)} \\ y &= 2x + 5 \quad \text{(Equation 2)} \end{aligned} \][/tex]
Step 1: Substitute Equation 2 into Equation 1
From Equation 2, we have:
[tex]\[ y = 2x + 5 \][/tex]
We can substitute this expression for \( y \) into Equation 1:
[tex]\[ 5x + 4(2x + 5) = 7 \][/tex]
Step 2: Simplify and solve for \( x \)
Distribute the \( 4 \) inside the parentheses:
[tex]\[ 5x + 8x + 20 = 7 \][/tex]
Combine the \( x \) terms:
[tex]\[ 13x + 20 = 7 \][/tex]
Subtract 20 from both sides to isolate the \( x \) term:
[tex]\[ 13x = 7 - 20 \][/tex]
Simplify the right side:
[tex]\[ 13x = -13 \][/tex]
Divide both sides by 13:
[tex]\[ x = \frac{-13}{13} \][/tex]
So,
[tex]\[ x = -1 \][/tex]
Step 3: Substitute \( x = -1 \) back into Equation 2 to solve for \( y \)
We use Equation 2 to find the corresponding \( y \) value:
[tex]\[ y = 2(-1) + 5 \][/tex]
Simplify:
[tex]\[ y = -2 + 5 \][/tex]
Therefore,
[tex]\[ y = 3 \][/tex]
Solution
The solution to the system of equations is:
[tex]\[ \begin{array}{l} x = -1 \\ y = 3 \end{array} \][/tex]
So, the point [tex]\((x, y) = (-1, 3)\)[/tex] satisfies both equations in the given system.
[tex]\[ \begin{aligned} 5x + 4y &= 7 \quad \text{(Equation 1)} \\ y &= 2x + 5 \quad \text{(Equation 2)} \end{aligned} \][/tex]
Step 1: Substitute Equation 2 into Equation 1
From Equation 2, we have:
[tex]\[ y = 2x + 5 \][/tex]
We can substitute this expression for \( y \) into Equation 1:
[tex]\[ 5x + 4(2x + 5) = 7 \][/tex]
Step 2: Simplify and solve for \( x \)
Distribute the \( 4 \) inside the parentheses:
[tex]\[ 5x + 8x + 20 = 7 \][/tex]
Combine the \( x \) terms:
[tex]\[ 13x + 20 = 7 \][/tex]
Subtract 20 from both sides to isolate the \( x \) term:
[tex]\[ 13x = 7 - 20 \][/tex]
Simplify the right side:
[tex]\[ 13x = -13 \][/tex]
Divide both sides by 13:
[tex]\[ x = \frac{-13}{13} \][/tex]
So,
[tex]\[ x = -1 \][/tex]
Step 3: Substitute \( x = -1 \) back into Equation 2 to solve for \( y \)
We use Equation 2 to find the corresponding \( y \) value:
[tex]\[ y = 2(-1) + 5 \][/tex]
Simplify:
[tex]\[ y = -2 + 5 \][/tex]
Therefore,
[tex]\[ y = 3 \][/tex]
Solution
The solution to the system of equations is:
[tex]\[ \begin{array}{l} x = -1 \\ y = 3 \end{array} \][/tex]
So, the point [tex]\((x, y) = (-1, 3)\)[/tex] satisfies both equations in the given system.
