Answer :
To solve the system of equations:
Equation 1: \(5x - 2y = -11\)
Equation 2: \(-2x + 5y = 17\)
we can create \(x\)-coefficients that are additive inverses (i.e., coefficients that sum to zero). To achieve this, follow these steps:
1. Choose suitable multipliers to make the coefficients of \(x\) in both equations additive inverses.
Step 1: To create \(x\)-coefficients that are additive inverses, Equation 1 can be multiplied by:
[tex]\[ \boxed{2} \][/tex]
Equation 1 becomes:
[tex]\[ 2 \cdot (5x - 2y) = 2 \cdot (-11) \][/tex]
[tex]\[ 10x - 4y = -22 \][/tex]
Step 2: Equation 2 can be multiplied by:
[tex]\[ \boxed{5} \][/tex]
Equation 2 becomes:
[tex]\[ 5 \cdot (-2x + 5y) = 5 \cdot 17 \][/tex]
[tex]\[ -10x + 25y = 85 \][/tex]
Now we have the modified system:
[tex]\[ 10x - 4y = -22 \][/tex]
[tex]\[ -10x + 25y = 85 \][/tex]
Here, we can see that the \(x\)-coefficients (\(10x\) and \(-10x\)) are additive inverses. Adding these two equations together will eliminate the \(x\) variable:
[tex]\[ (10x - 10x) + (-4y + 25y) = -22 + 85 \][/tex]
[tex]\[ 0 + 21y = 63 \][/tex]
[tex]\[ 21y = 63 \][/tex]
[tex]\[ y = 3 \][/tex]
Now, substitute \(y = 3\) back into one of the original equations to solve for \(x\). Using Equation 1:
[tex]\[ 5x - 2(3) = -11 \][/tex]
[tex]\[ 5x - 6 = -11 \][/tex]
[tex]\[ 5x = -11 + 6 \][/tex]
[tex]\[ 5x = -5 \][/tex]
[tex]\[ x = -1 \][/tex]
The solution to the system of equations is:
[tex]\[ \boxed{(-1, 3)} \][/tex]
Equation 1: \(5x - 2y = -11\)
Equation 2: \(-2x + 5y = 17\)
we can create \(x\)-coefficients that are additive inverses (i.e., coefficients that sum to zero). To achieve this, follow these steps:
1. Choose suitable multipliers to make the coefficients of \(x\) in both equations additive inverses.
Step 1: To create \(x\)-coefficients that are additive inverses, Equation 1 can be multiplied by:
[tex]\[ \boxed{2} \][/tex]
Equation 1 becomes:
[tex]\[ 2 \cdot (5x - 2y) = 2 \cdot (-11) \][/tex]
[tex]\[ 10x - 4y = -22 \][/tex]
Step 2: Equation 2 can be multiplied by:
[tex]\[ \boxed{5} \][/tex]
Equation 2 becomes:
[tex]\[ 5 \cdot (-2x + 5y) = 5 \cdot 17 \][/tex]
[tex]\[ -10x + 25y = 85 \][/tex]
Now we have the modified system:
[tex]\[ 10x - 4y = -22 \][/tex]
[tex]\[ -10x + 25y = 85 \][/tex]
Here, we can see that the \(x\)-coefficients (\(10x\) and \(-10x\)) are additive inverses. Adding these two equations together will eliminate the \(x\) variable:
[tex]\[ (10x - 10x) + (-4y + 25y) = -22 + 85 \][/tex]
[tex]\[ 0 + 21y = 63 \][/tex]
[tex]\[ 21y = 63 \][/tex]
[tex]\[ y = 3 \][/tex]
Now, substitute \(y = 3\) back into one of the original equations to solve for \(x\). Using Equation 1:
[tex]\[ 5x - 2(3) = -11 \][/tex]
[tex]\[ 5x - 6 = -11 \][/tex]
[tex]\[ 5x = -11 + 6 \][/tex]
[tex]\[ 5x = -5 \][/tex]
[tex]\[ x = -1 \][/tex]
The solution to the system of equations is:
[tex]\[ \boxed{(-1, 3)} \][/tex]
