Answer :
To solve the given system of linear equations using substitution, let's approach each set of equations step-by-step.
System 1:
1. [tex]\( y = -2x - 9 \)[/tex]
2. [tex]\( 6x - 5y = -19 \)[/tex]
Step 1: Substitute [tex]\( y \)[/tex] from the first equation into the second equation.
[tex]\[ 6x - 5(-2x - 9) = -19 \][/tex]
Step 2: Simplify and solve for [tex]\( x \)[/tex].
[tex]\[ 6x + 10x + 45 = -19 \][/tex]
[tex]\[ 16x + 45 = -19 \][/tex]
[tex]\[ 16x = -64 \][/tex]
[tex]\[ x = -4 \][/tex]
Step 3: Substitute [tex]\( x = -4 \)[/tex] back into the first equation to find [tex]\( y \)[/tex].
[tex]\[ y = -2(-4) - 9 \][/tex]
[tex]\[ y = 8 - 9 \][/tex]
[tex]\[ y = -1 \][/tex]
So, the solution to the first system is [tex]\( (x, y) = (-4, -1) \)[/tex].
System 2:
1. [tex]\( x = 6y - 7 \)[/tex]
2. [tex]\( 4x + y = -3 \)[/tex]
Step 1: Substitute [tex]\( x \)[/tex] from the first equation into the second equation.
[tex]\[ 4(6y - 7) + y = -3 \][/tex]
Step 2: Simplify and solve for [tex]\( y \)[/tex].
[tex]\[ 24y - 28 + y = -3 \][/tex]
[tex]\[ 25y - 28 = -3 \][/tex]
[tex]\[ 25y = 25 \][/tex]
[tex]\[ y = 1 \][/tex]
Step 3: Substitute [tex]\( y = 1 \)[/tex] back into the first equation to find [tex]\( x \)[/tex].
[tex]\[ x = 6(1) - 7 \][/tex]
[tex]\[ x = 6 - 7 \][/tex]
[tex]\[ x = -1 \][/tex]
So, the solution to the second system is [tex]\( (x, y) = (-1, 1) \)[/tex].
Therefore, the solutions are:
1. [tex]\( (x, y) = (-4, -1) \)[/tex]
2. [tex]\( (x, y) = (-1, 1) \)[/tex]
System 1:
1. [tex]\( y = -2x - 9 \)[/tex]
2. [tex]\( 6x - 5y = -19 \)[/tex]
Step 1: Substitute [tex]\( y \)[/tex] from the first equation into the second equation.
[tex]\[ 6x - 5(-2x - 9) = -19 \][/tex]
Step 2: Simplify and solve for [tex]\( x \)[/tex].
[tex]\[ 6x + 10x + 45 = -19 \][/tex]
[tex]\[ 16x + 45 = -19 \][/tex]
[tex]\[ 16x = -64 \][/tex]
[tex]\[ x = -4 \][/tex]
Step 3: Substitute [tex]\( x = -4 \)[/tex] back into the first equation to find [tex]\( y \)[/tex].
[tex]\[ y = -2(-4) - 9 \][/tex]
[tex]\[ y = 8 - 9 \][/tex]
[tex]\[ y = -1 \][/tex]
So, the solution to the first system is [tex]\( (x, y) = (-4, -1) \)[/tex].
System 2:
1. [tex]\( x = 6y - 7 \)[/tex]
2. [tex]\( 4x + y = -3 \)[/tex]
Step 1: Substitute [tex]\( x \)[/tex] from the first equation into the second equation.
[tex]\[ 4(6y - 7) + y = -3 \][/tex]
Step 2: Simplify and solve for [tex]\( y \)[/tex].
[tex]\[ 24y - 28 + y = -3 \][/tex]
[tex]\[ 25y - 28 = -3 \][/tex]
[tex]\[ 25y = 25 \][/tex]
[tex]\[ y = 1 \][/tex]
Step 3: Substitute [tex]\( y = 1 \)[/tex] back into the first equation to find [tex]\( x \)[/tex].
[tex]\[ x = 6(1) - 7 \][/tex]
[tex]\[ x = 6 - 7 \][/tex]
[tex]\[ x = -1 \][/tex]
So, the solution to the second system is [tex]\( (x, y) = (-1, 1) \)[/tex].
Therefore, the solutions are:
1. [tex]\( (x, y) = (-4, -1) \)[/tex]
2. [tex]\( (x, y) = (-1, 1) \)[/tex]
