Use the substitution method to solve the system 2x+3y=4 , −4x−6y=−8 .
no solution no solution (2−1) left parenthesis 2 comma negative 1 right parenthesis infinite solutions infinite solutions (−1,2)
To solve the system using the substitution method, we first solve one of the equations for one variable in terms of the other variable, and then substitute this expression into the other equation.
From the first equation, we can solve for x in terms of y:
2x + 3y = 4
2x = 4 - 3y
x = (4 - 3y) / 2
Now, we substitute this expression for x into the second equation:
-4x - 6y = -8
-4((4 - 3y) / 2) - 6y = -8
-2(4 - 3y) - 6y = -8
-8 + 6y - 6y = -8
-8 = -8
This equation simplifies to -8 = -8. Since this equation is always true, it implies that the system has infinite solutions.
Therefore, the correct answer is: infinite solutions.
Use the substitution method to solve the system x=y−4, x+8y=2.
(−313,23) left parenthesis negative 3 Start Fraction 1 over 3 End Fraction comma Start Fraction 2 over 3 End Fraction right parenthesis no solution no solution infinite solutions infinite solutions (23,−313)
To solve the system using the substitution method, we first solve one of the equations for one variable in terms of the other variable, and then substitute this expression into the other equation.
From the first equation, we can solve for x in terms of y:
x = y - 4
Now, we substitute this expression for x into the second equation:
(y - 4) + 8y = 2
9y - 4 = 2
9y = 6
y = 6/9
y = 2/3
Now that we have found the value of y, we can substitute it back into the first equation to find x:
x = (2/3) - 4
x = -12/3 - 4
x = -4 - 4
x = -8
Therefore, the solution to the system of equations is (-8, 2/3), which can also be written as (-313, 23) in fraction form.
The correct answer is: (-313, 23).
Use the substitution method to solve the system x=y−4, x+8y=2.
(−3 1/3,2/3) left parenthesis negative 3 Start Fraction 1 over 3 End Fraction comma Start Fraction 2 over 3 End Fraction right parenthesis no solution no solution infinite solutions infinite solutions (2/3,−3 1/3)
To solve the system using the substitution method, we first solve one of the equations for one variable in terms of the other variable, and then substitute this expression into the other equation.
From the first equation, we can solve for x in terms of y:
x = y - 4
Now, we substitute this expression for x into the second equation:
(y - 4) + 8y = 2
9y - 4 = 2
9y = 6
y = 6/9
y = 2/3
Now that we have found the value of y, we can substitute it back into the first equation to find x:
x = (2/3) - 4
x = -12/3 - 4
x = -4 - 4
x = -8
Therefore, the solution to the system of equations is (-8, 2/3), which can also be written as (-3 1/3, 2/3) in mixed fraction form.
The correct answer is: (-3 1/3, 2/3).
Use the substitution method to solve the system 2.5x+y=−2 , 3x+2y=0 .
no solution no solution (−2,3) left parenthesis negative 2 comma 3 right parenthesis infinite solutions infinite solutions (3,−2) left parenthesis 3 comma negative 2 right parenthesis
To solve the system using the substitution method, we first solve one of the equations for one variable in terms of the other variable, and then substitute this expression into the other equation.
From the first equation, we can solve for y in terms of x:
y = -2.5x - 2
Now, we substitute this expression for y into the second equation:
3x + 2(-2.5x - 2) = 0
3x - 5x - 4 = 0
-2x - 4 = 0
-2x = 4
x = 4/(-2)
x = -2
Now that we have found the value of x, we can substitute it back into the first equation to find y:
y = -2.5(-2) - 2
y = 5 - 2
y = 3
Therefore, the solution to the system of equations is (-2, 3).
The correct answer is: (-2, 3).
Which of the followin(4,8) and (3,5) left parenthesis 4 comma 8 right parenthesis and left parenthesis 3 comma 5 right parenthesis (4,8) left parenthesis 4 comma 8 right parenthesis (3,5) left parenthesis 3 comma 5 right parenthesis (−4,−16) and (−3,−13)g uses the substitution method to solve linear-quadratic system y=x^2−4x+8 , y=3x−4 ?(
To determine which solution uses the substitution method, we need to substitute one equation into the other and check if the coordinates satisfy both equations.
Let's substitute y = 3x - 4 into the equation y = x^2 - 4x + 8:
x^2 - 4x + 8 = 3x - 4
Rearranging the equation:
x^2 - 4x - 3x + 8 + 4 = 0
x^2 - 7x + 12 = 0
Factoring the quadratic equation:
(x - 3)(x - 4) = 0
This means that x = 3 or x = 4.
Now, let's substitute these x-values back into y = 3x - 4:
For x = 3:
y = 3(3) - 4
y = 9 - 4
y = 5
For x = 4:
y = 3(4) - 4
y = 12 - 4
y = 8
The two coordinates that satisfy both equations are (3, 5) and (4, 8).
Therefore, the correct answer is: (3, 5) and (4, 8).
To solve the system of equations using the substitution method, we need to solve one equation for one variable and substitute it into the other equation. Let's start by solving the first equation for x.
Given:
Equation 1: 2x + 3y = 4
Equation 2: -4x - 6y = -8
We'll begin by solving Equation 1 for x:
2x = 4 - 3y
x = (4 - 3y) / 2
Now we'll substitute this expression for x into Equation 2:
-4((4 - 3y) / 2) - 6y = -8
Simplifying Equation 2:
-2(4 - 3y) - 6y = -8
-8 + 6y - 6y = -8
-8 = -8
The equation simplifies to -8 = -8, which means that the equation is always true, regardless of the value of y. This indicates that the two equation lines are actually representing the same line, thus having infinitely many solutions.
Therefore, the system of equations has infinite solutions.