1
If \(7+h=3m-n\), which equation expresses \(h\) in terms of \(m\) and \(n\)?
Nonlinear equations in one variable and systems of equations in two variables: worksheet with answers
Work through the questions first, then open the answer key and explanations to check your work.
2
The graph shows the system of equations \(y=|x+2|-1\) and \(y=x+5\).
Which ordered pair is the solution to the system?
The coordinate plane has labeled axes from -8 to 8 on both axes with grid lines at each integer value.
3
The graph shows the line \(y=-x+2\) and the absolute value function \(y=|x+2|-1\). What is the ordered-pair solution to the system represented by the graph?
The coordinate plane includes labeled axes from -6 to 6 on both axes with a grid at each integer value.
4
If \(15uv=w\), where \(u\), \(v\), and \(w\) are positive numbers, which expression represents \(v\) in terms of \(u\) and \(w\)?
5
If \(q=-\frac{18r}{s}\), where \(s\ne 0\), which equation correctly expresses \(r\) in terms of \(q\) and \(s\)?
6
For real numbers \(m\), \(n\), and \(t\), where \(t\ne 0\), suppose the equation below is true.
\(\frac{n}{t}=5m-7\)
Which choice expresses \(n\) in terms of \(m\) and \(t\)?
7
For all real numbers \(m\), \(n\), and \(t\) such that \(t\ne 0\), the equation \(\frac{n}{t}=5m-8\) holds. Which choice expresses \(n\) in terms of \(m\) and \(t\)?
8
If \(18n^{2}=1458\), what is the positive solution for \(n\)?
9
Let \(x\), \(y\), and \(z\) be positive numbers with \(5y>2z\). If
\(\frac{10}{x}=\frac{5}{y}-\frac{2}{z}\),
what is the value of \(x\) in terms of \(y\) and \(z\)?
10
What is the sum of all real solutions to the equation \(|6y-5|=19\)?
11
What is the sum of all real solutions to the equation \(|6y-5|=13\)?
12
For positive numbers \(p\), \(q\), and \(r\) with \(q>2r\), suppose the following equation is true:
\(\frac{12}{p}-\frac{4}{q}=\frac{8}{r}\)
Which expression is equivalent to \(p\) in terms of \(q\) and \(r\)?
Answer key and explanations
1. Answer
\(h=3m-n-7\)
To isolate \(h\) in the equation \(7+h=3m-n\), subtract 7 from both sides. This gives \(h=3m-n-7\). Therefore, the correct choice is C.
2. Answer
\((-4,1)\)
The solution to a system of equations is the point where the graphs intersect. On the graph, the line and the absolute value graph meet at the point \((-4,1)\). Therefore, the solution to the system is \((-4,1)\).
Choice A is the vertex of the absolute value graph, choice B is the y-intercept of the line, and choice D reflects a sign error.
3. Answer
\((\frac{1}{2},\frac{3}{2})\)
The solution to a system of equations is the point where the graphs intersect. On the graph, the line and the V-shaped absolute value graph meet at the point \((\frac{1}{2},\frac{3}{2})\). Therefore, the ordered-pair solution is \((\frac{1}{2},\frac{3}{2})\).
The point \((-2,-1)\) is the vertex of the absolute value graph, not the intersection. The point \((0,2)\) is the y-intercept of the line. The point \((\frac{3}{2},\frac{1}{2})\) reverses the coordinates of the correct answer.
4. Answer
\(v=\frac{w}{15u}\)
Starting with \(15uv=w\), isolate \(v\) by dividing both sides of the equation by \(15u\). This gives \(v=\frac{w}{15u}\).
Choice A subtracts instead of dividing. Choice B multiplies by \(15u\) instead of dividing. Choice D incorrectly changes the multiplicative relationship into an additive one.
5. Answer
\(r=-\frac{qs}{18}\)
Start with \(q=-\frac{18r}{s}\). Multiply both sides by \(s\) to clear the denominator:
\(qs=-18r\)
Now divide both sides by \(-18\):
\(r=-\frac{qs}{18}\)
Therefore, the correct answer is choice C.
6. Answer
\(n=5mt-7t\)
Start with the equation \(\frac{n}{t}=5m-7\). Multiply both sides by \(t\) to clear the denominator:
\(n=t(5m-7)\)
Then distribute \(t\) across the parentheses:
\(n=5mt-7t\)
Therefore, the correct choice is C.
7. Answer
\(n=5mt-8t\)
Start with the equation \(\frac{n}{t}=5m-8\). Multiply both sides by \(t\) to clear the denominator:
\(n=t(5m-8)\)
Now distribute \(t\) across the parentheses:
\(n=5mt-8t\)
Therefore, the correct choice is the expression \(5mt-8t\).
8. Answer
9
Divide both sides of the equation by 18 to isolate \(n^{2}\):
\(n^{2}=\frac{1458}{18}=81\)
Taking the square root of both sides gives \(n=\pm 9\). Because the question asks for the positive solution, the answer is \(9\).
9. Answer
\(\frac{10yz}{5z-2y}\)
Start by combining the fractions on the right side using the common denominator \(yz\):
\(\frac{5}{y}-\frac{2}{z}=\frac{5z-2y}{yz}\).
So the equation becomes
\(\frac{10}{x}=\frac{5z-2y}{yz}\).
Multiply both sides by \(x\) and by \(yz\):
\(10yz=x(5z-2y)\).
Now solve for \(x\):
\(x=\frac{10yz}{5z-2y}\).
Therefore, the correct answer is choice C.
10. Answer
\(\frac{5}{3}\)
To solve \(|6y-5|=19\), split the equation into two cases:
\(6y-5=19\) or \(6y-5=-19\).
For the first equation:
\(6y=24\), so \(y=4\).
For the second equation:
\(6y=-14\), so \(y=-\frac{7}{3}\).
The sum of the solutions is
\(4+\left(-\frac{7}{3}\right)=\frac{12}{3}-\frac{7}{3}=\frac{5}{3}\).
Therefore, the correct answer is \(\frac{5}{3}\).
11. Answer
\(\frac{5}{3}\)
To solve an absolute value equation, split it into two cases.
\(6y-5=13\) gives \(6y=18\), so \(y=3\).
\(6y-5=-13\) gives \(6y=-8\), so \(y=-\frac{4}{3}\).
The sum of the solutions is \(3+\left(-\frac{4}{3}\right)=\frac{9}{3}-\frac{4}{3}=\frac{5}{3}\).
Therefore, the correct answer is \(\frac{5}{3}\).
12. Answer
\(\frac{3qr}{2q+r}\)
Start with the given equation:
\(\frac{12}{p}-\frac{4}{q}=\frac{8}{r}\)
Add \(\frac{4}{q}\) to both sides:
\(\frac{12}{p}=\frac{8}{r}+\frac{4}{q}\)
Find a common denominator on the right side:
\(\frac{8}{r}+\frac{4}{q}=\frac{8q+4r}{qr}\)
Factor the numerator:
\(\frac{8q+4r}{qr}=\frac{4(2q+r)}{qr}\)
So,
\(\frac{12}{p}=\frac{4(2q+r)}{qr}\)
Divide both sides by 4:
\(\frac{3}{p}=\frac{2q+r}{qr}\)
Now take reciprocals:
\(\frac{p}{3}=\frac{qr}{2q+r}\)
Multiply by 3:
\(p=\frac{3qr}{2q+r}\)
Therefore, the correct answer is choice C.