The question asks for the value of m where costs are equal: 15 + 0.10m = 5 + 0.20m. I will use the substitution method by isolating terms. Subtract 0.10m: 15 = 5 + 0.10m; subtract 5: 10 = 0.10m, m = 100. A common error is subtracting 0.10m from the wrong side, leading to m = 50. Verify by calculating costs at m = 100: both equal 25 dollars. Graphing the lines can help visualize the intersection.

The question asks for the ordered pair (m, c) where the two plans cost the same. I will use the substitution method by setting the cost expressions equal. So 6 + 2m = 12 + m, 2m - m = 12 - 6, m = 6. Then c = 12 + 6 = 18. The pair is (6, 18). A common error is forgetting the base fees or mis-setting the equations. It is easy to solve for m but not check the total cost. As a strategy, interpret the intersection as the break-even point and verify.

The question requires finding the intersection point (x, y) of lines ℓ1 through (-4,5) and (0,1), and ℓ2 through (0,-2) and (2,2). I will find the equations of the lines and then use substitution to solve the system. For ℓ1, the slope is (1-5)/(0-(-4)) = -1, so y = -x + 1; for ℓ2, the slope is (2-(-2))/(2-0) = 2, so y = 2x - 2. Set equal: -x + 1 = 2x - 2, so 3 = 3x and x = 1, then y = -1 + 1 = 0, giving (1,0). A common error is calculating the slope incorrectly by swapping points. Verify by plugging the point back into both line equations.

The question asks how many solutions the system {5x - 3y = 1, 10x - 6y = 4} has, considering if the lines are parallel, intersecting, or the same. I will use the elimination method by checking multiples. Multiplying the first equation by 2 gives 10x - 6y = 2, which has the same coefficients as the second but a different constant (2 ≠ 4). This indicates parallel lines that never intersect, so there are 0 solutions. A key error is overlooking the constant difference after scaling, which might lead to thinking there are infinitely many solutions.

We need to determine if this system has no solutions, one solution, or infinitely many solutions. Let me examine the relationship between the two equations. The second equation is 2x - y = 5. If I multiply this by 2, I get 4x - 2y = 10, which is exactly the first equation. Since the first equation is just a multiple of the second equation, they represent the same line. When two equations represent the same line, every point on that line is a solution, giving us infinitely many solutions. A common error would be to think these are parallel lines because the coefficients are proportional, but since the constants are also proportional (10/5 = 2), they're the same line. For test-taking, when you see one equation is a multiple of another, immediately check if it's the exact same line (infinitely many solutions) or parallel lines (no solutions).

We need to determine how many solutions this system has. First, I'll rewrite the second equation in slope-intercept form: x - 2y = 8 becomes -2y = -x + 8, so y = (1/2)x - 4. Now we have y = (1/2)x + 3 and y = (1/2)x - 4. Both lines have the same slope (1/2) but different y-intercepts (3 and -4), which means they are parallel lines that never intersect. Parallel lines have no solutions. The common error would be to think these represent the same line without checking the y-intercepts. For test-taking, when both equations are in slope-intercept form, quickly compare slopes and y-intercepts: same slope with different intercepts means no solutions.

We need to determine how many solutions this system has: 6x + 9y = 12 and 2x + 3y = 5. Let me check if one equation is a multiple of the other. If I multiply the second equation by 3: 6x + 9y = 15. Comparing with the first equation 6x + 9y = 12, we see the left sides are identical but the right sides are different (15 ≠ 12). This means the lines are parallel (same slope) but have different y-intercepts, so they never intersect. Therefore, there are 0 solutions. The warning is correct - just because coefficients are proportional doesn't mean the equations represent the same line; you must check the constants too. For test-taking, when coefficients are proportional, immediately check if constants are also proportional.

We need to find where y = -x + 2 and y = 2x - 7 intersect. Since both equations are solved for y, I'll set them equal: -x + 2 = 2x - 7. Adding x to both sides: 2 = 3x - 7. Adding 7 to both sides: 9 = 3x, so x = 3. Substituting back into the first equation: y = -3 + 2 = -1. Therefore the solution is (3, -1), which is answer C. The warning about forgetting constants is important - if you just set -x = 2x, you'd get x = 0, missing the constant terms entirely. For test-taking, when both equations are in y = form, setting them equal is the most efficient method.

We need to find the point that satisfies both y = 2x - 7 and 3x + y = 8. I'll use substitution since y is already isolated in the first equation. Substituting y = 2x - 7 into the second equation: 3x + (2x - 7) = 8, which simplifies to 5x - 7 = 8, so 5x = 15, and x = 3. Now finding y: y = 2(3) - 7 = 6 - 7 = -1. Therefore the solution is (3, -1), which matches answer A. The warning about substituting for x instead of y is important - if you mistakenly replace x with 2x - 7, you'd get 3(2x - 7) + y = 8, leading to an incorrect solution. When using substitution with equations in the form y = expression, always substitute that expression for y, not for x.

We need to find the per-mile rate y, where x is the flat fee, using the information that a 6-mile trip costs $17 and a 10-mile trip costs $25. I'll set up the system: x + 6y = 17 and x + 10y = 25. Using elimination by subtracting the first equation from the second: (x + 10y) - (x + 6y) = 25 - 17, which gives 4y = 8, so y = 2. This means the per-mile rate is $2. A common error would be confusing which variable represents what or setting up the equations incorrectly. When dealing with word problems, always define your variables clearly and check that your answer makes sense in context.