Solve algebra questions involving the following topics:
Solve the equation \[ 5(- 3x - 2) - (x - 3) = -4(4x + 5) + 13 \]
iven the equation \[ 5 (- 3 x - 2) - (x - 3) = - 4 (4 x + 5) + 13 \] Expand \[ -15 x - 10 - x + 3 = - 16 x - 20 + 13 \] Group like terms \[ - 16 x - 7 = - 16 x - 7 \] Add \( 16 x + 7 \) to both sides \[ - 16 x - 7 16 x + 7 = - 16 x - 7 16 x + 7 \] Simplify to obtain \[ 0 = 0 \]
The above statement is true for all values of \( x \) and therefore all real numbers are solutions to the given equation. This equation is called an identity
Simplify the expression \[ 2(a -3) + 4b - 2(a -b -3) + 5\]
Given the algebraic expression \[ 2 (a -3) + 4 b - 2 (a - b - 3) + 5 \] Expand the factors \[ = 2 a - 6 + 4 b - 2 a + 2 b + 6 + 5 \] Group like terms. \[ = 6 b + 5 \]
If \( x \lt 2 \), simplify the expression \[ |x - 2| - 4|-6| \]
Given the expression \[ | x - 2 | - 4 | -6 | \] If \( x \lt 2 \) then \[ x - 2 \lt 0 \]. According to the definition of the absolute value and since \( (x - 2) \lt 0 \) \[ |x - 2| = - (x - 2) \] and also \[ | - 6 | = 6 \] Substitute \( |x - 2| \) by \( - (x - 2) \) and \( | - 6 | \) by \( 6 \) in the given expression \[ |x - 2| - 4| -6 | = - (x - 2) - 4(6) = - x + 2 - 24\] Group like terms \[ |x - 2| - 4| -6 | = - x - 22 \]
Find the distance between the points (-4 , -5) and (-1 , -1).
According to the formula of the distance \( d \) between two points \( (x_1 , y_1) \) and \( (x_2 , y_2) \) \[ d = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2 } \] the distance \(d\) between the points \( (-4 , -5) \) and \( (-1 , -1) \) is given by (substitution) \[ d = \sqrt{ (-1 - (-4))^2 + (-1 - (-5))^2 } \] Simplify \[ d = \sqrt{ (-5)^2 + (4)^2 } \] \[ d = \sqrt{9 + 16} = 5 \]
Find the x intercept of the graph of the equation . \[ 2x - 4y = 9 \]
Given the equation \[ 2 x - 4 y = 9 \] To find the \( x \) intercept, we set \( y = 0 \) in the equation and solve for \( x \). \[ 2 x - 0 = 9 \] Solve for x \[ x = 9 / 2 \] The \( x \) intercept is at the point \( (9/2 , 0) \).
Evaluate \( f(2) - f(1) \) given \[ f(x) = 6x + 1 \]
Given the function \[ f(x) = 6 x + 1 \] \( f(2) \) is found by substituting \( x \) by \( 2 \) in \( f(x) \) and \( f(1) \) is found by substituting \( x \) by \( 1 \) in \( f(x) \); hence \[ f(2) - f(1) = (6 \times 2 + 1) - (6 \times 1 + 1) = 13 - 7 = 6 \]
Find the slope of the line passing through the points \( (-1, -1) \) and \( (2 , 2) \).
According to the formula of the slope of the line through the points \( (x_1 , y_1) \) and \( (x_2 , y_2) \) given by \[ m = \dfrac{y_2 - y_1}{x_2 - x_1} , \] the slope of the line through the points \( (-1, -1) \) and \( (2 , 2) \) is given by \[ m = \dfrac {2 - (-1)}{2 - (-1)} = \dfrac{2+1}{2+1} = 1 \]
Find the slope of the line \[ 5x - 10y = 7 \]
Given the equation of the line \[ 5x - 10y = 7 \] Rewrite the equation in slope intercept form \( y = m x + b \) \[ - 10 y = - 5 x + 7 \] Divide all terms by \( -10 \) \[ \dfrac{- 10 y} {-10} = \dfrac{- 5 x}{-10} + \dfrac{7}{-10} \] Simplify \[ y = \dfrac{1}{2} x - \dfrac{7}{10} \] The slope is given by the coefficient of \( x \) which is \( \dfrac{1}{2} \).
Find the equation of the line that passes through the points (-1 , -1) and (-1 , 2).
To find the equation of the line through the points \( (-1 , -1) \) and \( (-1 , 2) \), we first find the slope \( m \). \[ m = \dfrac{y_2 - y_1} {x_2 - x_1} = \dfrac{2 - (-1)}{-1 - (-1)} = \dfrac{2 +1}{-1 + 1} = \dfrac{3}{0} = \text{undefined} \] The divison by zero is undefined in mathematics and therefore the slope is undefined which means the line is perpendicular to the \( x \) axis and its equation has the form \( x = constant \).
Since both points have equal \( x \) coordinates \( -1 \), the equation of the line through the points \( (-1 , -1) \) and \( (-1 , 2) \) is a vertical given by: \[ x = -1 \]
Solve the equation \[ |- 2 x + 2| - 3 = -3 \]
The equation to solve is given by. \[ |-2 x + 2| -3 = -3 \] Add 3 to both sides of the equation \[ |-2 x + 2| -3 + 3 = -3 + 3 \] Group like terms \[ |-2 x + 2| = 0 \] According to the definition of the absolute value, \( |-2 x + 2| = 0 \) means \[ -2 x + 2 = 0 \] Solve for \( x \) to obtain \[ x = 1 \]