Question 1

Solve 4x + 2 = 0&#10;A) x = -1&#10;B) x = -0.5&#10;C) x = 0&#10;D) x = 0.5

Accepted Answer

x = 0.5

Question 2

Which of the equations below are NOT equivalent to x = 4?&#10;A) 2x = 8&#10;B) = 2&#10;C) x + 3 = 4 - x&#10;D) 3x - 6 = 6

Accepted Answer

x + 3 = 4 - x is not equivalent to x = 4 because it is not a true statement. When x = 4, x + 3 = 7, not 4.

Question 3

Solve  &#10;A) 3&#10;B) 4&#10;C) 5&#10;D) 6

Accepted Answer

5&#10;

Question 4

Find a solution for y when 2x - 2y = -6 and 2x + 4y = 30&#10;A) 6&#10;B) -6&#10;C) 3&#10;D) -3

Accepted Answer

To solve for y, we can eliminate x by adding the two equations together: (2x - 2y) + (2x + 4y) = -6 + 30, which simplifies to 4x + 2y = 24. Dividing the entire equation by 2 gives 2x + y = 12. However, this approach was incorrect in eliminating x directly. Correctly, subtracting the first equation from the second gives: (2x + 4y) - (2x - 2y) = 30 - (-6), simplifying to 6y = 36, and thus y = 6.

Question 5

Solve this set of 3x3 simultaneous equations for the value of y: 2y + x - 3z = -5; -4y - 2x +3z = -2; y + 2x + 2z = 16&#10;A) 1&#10;B) 2&#10;C) 3&#10;D) 4

Accepted Answer

To solve for y, we can use either substitution or elimination method to solve the system of equations. First, we can simplify and rearrange the equations if necessary to make the process easier. Let's use the elimination method to eliminate variables systematically until we solve for y.Given equations:1) 2y + x - 3z = -52) -4y - 2x + 3z = -23) y + 2x + 2z = 16First, we can add equations 1 and 2 to eliminate z:(2y + x - 3z) + (-4y - 2x + 3z) = -5 + (-2)This simplifies to:-2y - x = -7  ... (Equation 4)Next, we can multiply equation 3 by 2 to make it easier to eliminate x with equation 4:2(y + 2x + 2z) = 2(16)2y + 4x + 4z = 32  ... (Equation 5)Now, we need to find another pair of equations to eliminate another variable. We can use equation 4 and equation 1 to eliminate x by multiplying equation 4 by 1 (to keep it the same) and equation 1 by 1 as well, then add them together:-2y - x + 2y + x - 3z = -7 - 5This simplifies to:-3z = -12z = 4Now, we can substitute z = 4 into any of the original equations to find x or y. Let's substitute into equation 3:y + 2x + 2(4) = 16y + 2x + 8 = 16y + 2x = 8  ... (Equation 6)Now, let's eliminate x using equations 4 and 6. We can add equation 4 to equation 6 (after adjusting signs if necessary):-2y - x + y + 2x = -7 + 8-y + x = 1Since we need to solve for y, let's go back to our simplified equations to find a direct relationship. We can substitute z = 4 into equation 1 or 2 or 3 to solve for y directly. Let's use equation 1 for simplicity:2y + x - 3(4) = -52y + x - 12 = -52y + x = 7Now, we need to solve for y directly. We can use the values of x and z we found (or would find by substituting z into one of the equations) to solve for y. However, without finding x explicitly, we can look at our process to find any mistake or oversight in directly solving for y.Given the complexity of solving this system in this format and the mistake in not directly solving for y, let's correct the approach:The correct approach involves solving the system of equations step by step, either by substitution or elimination. However, without going through the entire process correctly and directly solving for y, the explanation provided does not directly lead to the solution for y. The correct approach would involve either substituting the values of z and x found into one of the equations to solve for y or continuing the elimination process correctly to isolate y.Upon correctly solving the system using either substitution or elimination method, you would find that y = 2, which corresponds to choice B. The explanation attempted to demonstrate the process but did not accurately follow through to the solution for y. The correct solution involves systematically eliminating variables and solving the resulting equations step by step.

Question 6

If a linear equation such as 4x + 2 = 8 is plotted graphically, which of the following features represents the solution?&#10;A) Where the line crosses the vertical (y) axis&#10;B) Where the line crosses the horizontal (x) axis&#10;C) The gradient of the line&#10;D) It is not possible to determine the solution in this way

Accepted Answer

The solution to a linear equation like 4x + 2 = 8, when plotted graphically, is represented by the point where the line crosses the horizontal (x) axis. This is because solving the equation for x gives the value of x when y is zero, which is exactly where the line crosses the x-axis.

Question 7

S &#10;A) 2&#10;B) 3&#10;C) 4&#10;D) 5

Accepted Answer

The answer of S &#10;A) 2&#10;B) 3&#10;C) 4&#10;D) 5...

Question 8

Find a solution for y when 6y + 6x = 66 and 2y = 6x - 34&#10;A) 2&#10;B) 4&#10;C) 6&#10;D) 8

Accepted Answer

To solve for y, we can use the second equation to express x in terms of y and then substitute that expression into the first equation. From the second equation, 2y = 6x - 34, we can solve for x: 6x = 2y + 34, so x = (2y + 34)/6. Substituting this into the first equation, 6y + 6((2y + 34)/6) = 66, simplifies to 6y + 2y + 34 = 66. Combining like terms gives 8y = 32, and dividing both sides by 8 gives y = 4.

Question 9

Which of the following equations does NOT have a single, unique, solution?&#10;A) 2x + 4 = 18&#10;B) = 6x - 3&#10;C) 4x + 2 = 3x + 6&#10;D) y = 2x - 4

Accepted Answer

The answer of Which of the following equations does NOT...

Question 10

When solving a 2x2 simultaneous equation graphically, which of the following solutions represents the solution?&#10;A) The difference between the points where the lines cross the vertical (y) axis&#10;B) The difference between the points where the lines cross the horizontal (x) axis&#10;C) The difference between the gradients of the two lines&#10;D) The point at which the two lines intersect (cross) each other

Accepted Answer

The solution to a 2x2 simultaneous equation graphically is represented by the point at which the two lines intersect each other. This point gives the values of x and y that satisfy both equations simultaneously.

Deck 25: Linear Equations