use the substitution method. Show your work. If the system has no solution or an infinite number of solutions, state this 3x + 24y = 24s,27x – 15y = -15 .
3x + 24y = 24
27x – 15y = -15
Divide first equation by 3.
x + 8y = 8
Transpose y term.
x = 8 - 8y
Substitute 8-8y for x in second equation and solve for y. Insert that value into the first equation and solve for x. Check by inserting both values into the second equation.
3x+24y=24
27x -15y=-15
solve for x in the first equation.
x= -y+8
put that in the second equation
27(-y+8)-15y=-15
solve for y
Then put that y back into the first equation, solve for x.
To solve a system of equations using the substitution method, follow these steps:
Step 1: Solve one equation for one variable.
Step 2: Substitute the expression obtained in Step 1 into the other equation.
Step 3: Solve the resulting equation for the remaining variable.
Step 4: Substitute the value found in Step 3 into one of the original equations to solve for the other variable.
Step 5: Check the solution in both original equations.
Now let's use the substitution method to solve the given system of equations:
Equation 1: 3x + 24y = 24s
Equation 2: 27x - 15y = -15
Step 1: Solve Equation 1 for x:
Rearrange Equation 1 to get x in terms of y and s:
3x + 24y = 24s
3x = 24s - 24y
x = 8s - 8y (Divide both sides of the equation by 3)
Step 2: Substitute the expression for x obtained in Step 1 into Equation 2:
27(8s - 8y) - 15y = -15
Step 3: Simplify the equation from Step 2 and solve for y:
216s - 216y - 15y = -15
216s - 231y = -15
-231y = -216s - 15
y = (216s + 15) / 231 (Divide both sides of the equation by -231)
Therefore, the value of y is (216s + 15) / 231.
Step 4: Substitute the value of y into Equation 1 to solve for x:
3x + 24((216s + 15) / 231) = 24s
3x + (8(216s + 15) / 3) = 24s
3x + (288s + 120) / 3 = 24s
3x + 96s + 40 = 72s
3x = -24s - 40
x = (-8s - 40) / 3 (Divide both sides of the equation by 3)
Therefore, the value of x is (-8s - 40) / 3.
Step 5: Check the solution in both original equations:
Equation 1: 3x + 24y = 24s
3((-8s - 40) / 3) + 24((216s + 15) / 231) = 24s
-8s - 40 + (192s + 120) / 231 = 24s
Equation 2: 27x - 15y = -15
27((-8s - 40) / 3) - 15((216s + 15) / 231) = -15
-8s - 40 - 5((192s + 120) / 231) = -15
Check both equations and simplify to see if the equations hold true. If they do, the solution is valid.
If the resulting equations in Step 5 continue to hold true for any value of s, then the system has an infinite number of solutions. If they don't hold true for any value of s, then the system has no solution. You can now proceed to perform the checks to determine the nature of the solutions in this particular case.