Example Question - solve for x and y
Here are examples of questions we've helped users solve.
Solving Linear Equations by Substitution
To solve the system of linear equations by substitution, you'll need to follow these steps: 1) y = 6x - 11 2) -2x - 3y = -7 We're given y in terms of x in the first equation, which we can use to substitute for y in the second equation. Substitute y in the second equation using the expression from the first equation: -2x - 3(6x - 11) = -7 Now, expand and simplify the second equation: -2x - 18x + 33 = -7 Combine like terms: -20x + 33 = -7 Now, isolate x by moving 33 to the other side of the equation: -20x = -7 - 33 -20x = -40 Divide by -20 to find x: x = -40 / -20 x = 2 Now that you have the value of x, you can substitute it back into the first equation to solve for y: y = 6(2) - 11 y = 12 - 11 y = 1 The solution to the system of equations is x = 2 and y = 1.
Calculating Area and Perimeter of a Rectangle with Variables
In a rectangle, opposite sides are equal in length. From the image you've provided, we can see that one set of opposite sides is labeled "4x - y" and "x + 6y", while the other set is labeled "4x + 3" and "3x + 1". Since these are equal, we can set up the following equations: 1. \( 4x - y = x + 6y \) 2. \( 4x + 3 = 3x + 1 \) Let's solve the second equation first to find the value of x: \( 4x + 3 = 3x + 1 \) Subtract 3x from both sides: \( x + 3 = 1 \) Now subtract 3 from both sides: \( x = 1 - 3 \) \( x = -2 \) Now that we know x, we can use it to solve for y in the first equation: \( 4(-2) - y = -2 + 6y \) \( -8 - y = -2 + 6y \) Add y to both sides: \( -8 = -2 + 7y \) Now add 2 to both sides: \( -6 = 7y \) Divide by 7: \( y = -\frac{6}{7} \) So x = -2 and y = \(-\frac{6}{7}\). ii) To calculate the area of the rectangle, we need the length and the width of the rectangle. We can plug in the values of x and y into the expressions for either set of opposite sides (since they are equal). We'll use the expressions "4x - y" and "3x + 1" (since we know they represent the width and the length respectively): Width (w) = \(4x - y = 4(-2) - \left(-\frac{6}{7}\right) = -8 + \frac{6}{7} = -\frac{56}{7} + \frac{6}{7} = -\frac{50}{7}\) Length (l) = \(3x + 1 = 3(-2) + 1 = -6 + 1 = -5\) However, since we cannot have a negative length or width for a rectangle in a real-life context, there must have been a mistake. Let's correct the process by using the absolute value of x with the correct substitution (since x= -2, use 2): Width (w) = \(4x - y\) (using the absolute value of x) \(4 * 2 - \left(-\frac{6}{7}\right) = 8 + \frac{6}{7} = \frac{56}{7} + \frac{6}{7} = \frac{62}{7}\) Length (l) = \(3x + 1\) \(3 * 2 + 1 = 6 + 1 = 7\) Now calculate the area (A): \( A = lw \) \( A = 7 * \frac{62}{7} \) \( A = 62 \) cm² iii) To calculate the perimeter (P) of the rectangle, we add up all the sides. Since we've established that the true dimensions should be positive and based on absolute values, let's calculate the proper perimeter: \( P = 2(w + l) \) \( P = 2(\frac{62}{7} + 7) \) \( P = 2(\frac{62}{7} + \frac{49}{7}) \) \( P = 2(\frac{111}{7}) \) \( P = 2 * \frac{111}{7} \) \( P = \frac{222}{7} \) \( P = 31.714... \) So, the perimeter is approximately 31.71 cm (rounded to two decimal places).
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