1
What is the value of \(x\) in the system of equations below?
\(y=2x-5\)
\(3x+y=15\)
Systems of two linear equations in two variables: worksheet with answers
Work through the questions first, then open the answer key and explanations to check your work.
2
If \(\begin{cases} 3m-2n=11 \\ m+2n=5 \end{cases}\), what is the value of \(2m-n\)?
3
The graph shows two linear equations on the same coordinate plane. What is the solution to the system of equations?
Graph description (alt text): The x-axis and y-axis each run from -6 to 6 with integer tick marks. One line rises from left to right and passes through the points (-4, -5) and (2, 1). The other line falls from left to right and passes through the points (-2, 5) and (2, 1). The two lines intersect at the point (2, 1).
4
The graph shows two lines on the coordinate plane. What is the solution to the system represented by these lines?
The red line passes through the points \((0,1)\) and \((4,5)\). The blue line passes through the points \((0,7)\) and \((6,1)\).
5
A sporting goods store sold a pair of hiking boots and a backpack for a combined original price of $180. During a weekend sale, the hiking boots were discounted by 25% and the backpack was discounted by 40%. After the discounts, the combined sale price was $123.
If \(b\) represents the original price of the hiking boots and \(p\) represents the original price of the backpack, which system of equations correctly models the situation?
6
If \(\begin{cases} 3m-2n=7 \\ m+n=5 \end{cases}\), what is the value of \(2m+n\)?
7
Consider the system of equations:
\(3x + (5y - 2) = 19\)
\(5y - 2 = 7\)
What is the value of \(x\)?
8
What is the solution to the following system of equations?
\(y=-3x+11\)
\(4x-2y=-6\)
9
The graph shows the lines \(y=-x+1\) and \(y=3\).
The two lines intersect at one point on the coordinate plane.
If the equation \(2x+y=8\) is added to form a system of three equations, how many solutions does the resulting system have?
10
The graph shows two linear equations. One line passes through the points \((-4,1)\) and \((4,5)\). The other line is the horizontal line \(y=3\).
The resulting system of three equations includes those two equations and the equation \(2x-y=3\).
How many solutions does the resulting system of three equations have?
11
For what value of \(p\) does the system of equations below have no solution?
\(6x-2(y-5)=20\)
\(4x+3y=px-9\)
12
For what value of \(p\) does the following system of equations have no solution?
\(4x-3y+8=2x+y-6\)
\(5x+2y=px-10\)
Answer key and explanations
1. Answer
4
Substitute \(2x-5\) for \(y\) in the second equation:
\(3x+(2x-5)=15\)
Combine like terms:
\(5x-5=15\)
Add 5 to both sides:
\(5x=20\)
Divide by 5:
\(x=4\)
Therefore, the correct answer is 4.
2. Answer
\(\frac{15}{2}\)
Add the two equations to eliminate \(n\):
\((3m-2n)+(m+2n)=11+5\)
\(4m=16\), so \(m=4\).
Substitute into \(m+2n=5\):
\(4+2n=5\)
\(2n=1\), so \(n=\frac{1}{2}\).
Now evaluate the requested expression:
\(2m-n=2(4)-\frac{1}{2}=8-\frac{1}{2}=\frac{15}{2}\).
Therefore, the correct answer is \(\frac{15}{2}\).
3. Answer
\((2,1)\)
The solution to a system of two linear equations is the point where the two lines intersect on the graph. Both lines cross at \((2,1)\), so that ordered pair is the solution.
4. Answer
\((3,4)\)
The solution to a system of linear equations is the point where the two lines intersect.
The red line through \((0,1)\) and \((4,5)\) has slope \(1\), so its equation is \(y=x+1\).
The blue line through \((0,7)\) and \((6,1)\) has slope \(-1\), so its equation is \(y=-x+7\).
At the intersection, both equations have the same \(x\) and \(y\) values:
\(x+1=-x+7\)
\(2x=6\), so \(x=3\).
Substitute into either equation: \(y=3+1=4\).
The solution is \((3,4)\).
5. Answer
\(\begin{cases}b+p=180\\0.75b+0.60p=123\end{cases}\)
The original prices add to 180, so the first equation is \(b+p=180\).
A 25% discount means the hiking boots retain 75% of their original price, so the discounted price is \(0.75b\). A 40% discount means the backpack retains 60% of its original price, so the discounted price is \(0.60p\). Since the combined sale price is 123, the second equation is \(0.75b+0.60p=123\).
Therefore, the correct system is choice C.
6. Answer
\(\frac{42}{5}\)
Use substitution or elimination to solve the system. From \(m+n=5\), we get \(n=5-m\). Substitute into the first equation:
\(3m-2(5-m)=7\)
\(3m-10+2m=7\)
\(5m=17\)
\(m=\frac{17}{5}\)
Then
\(n=5-\frac{17}{5}=\frac{8}{5}\).
Now evaluate the expression:
\(2m+n=2\left(\frac{17}{5}\right)+\frac{8}{5}=\frac{34}{5}+\frac{8}{5}=\frac{42}{5}\).
Therefore, the correct answer is \(\frac{42}{5}\).
7. Answer
4
The expression \(5y - 2\) appears in both equations. From the second equation, \(5y - 2 = 7\).
Substitute 7 for \(5y - 2\) in the first equation:
\(3x + 7 = 19\)
Subtract 7 from both sides:
\(3x = 12\)
Divide by 3:
\(x = 4\)
8. Answer
\(\left(\frac{8}{5},\frac{31}{5}\right)\)
Substitute the expression for \(y\) from the first equation into the second equation:
\(4x-2(-3x+11)=-6\)
Distribute the \(-2\):
\(4x+6x-22=-6\)
Combine like terms:
\(10x-22=-6\)
Add 22 to both sides:
\(10x=16\)
Divide by 10:
\(x=\frac{8}{5}\)
Now substitute \(x=\frac{8}{5}\) into \(y=-3x+11\):
\(y=-3\left(\frac{8}{5}\right)+11=-\frac{24}{5}+\frac{55}{5}=\frac{31}{5}\)
Therefore, the solution is \(\left(\frac{8}{5},\frac{31}{5}\right)\).
9. Answer
Zero solutions
The original two lines intersect where \(y=3\) and \(y=-x+1\). Substituting \(y=3\) into \(y=-x+1\) gives \(3=-x+1\), so \(x=-2\). The intersection point is \((-2,3)\).
Any solution to the system of three equations must satisfy all three equations, so it must be this intersection point. Check whether \((-2,3)\) satisfies \(2x+y=8\):
\(2(-2)+3=-4+3=-1\), which is not equal to 8.
Therefore, no point satisfies all three equations, so the system has zero solutions.
10. Answer
Zero solutions
The two original lines intersect exactly once. Since one line is \(y=3\), substitute \(y=3\) into the slanted line.
The slanted line passes through \((-4,1)\) and \((4,5)\), so its slope is \(\frac{5-1}{4-(-4)}=\frac{4}{8}=\frac{1}{2}\). Its equation is \(y=\frac{1}{2}x+3\). Setting \(y=3\) gives \(3=\frac{1}{2}x+3\), so \(x=0\). The intersection point is \((0,3)\).
Now test whether this point satisfies the third equation:
\(2(0)-3=-3\), which is not equal to \(3\).
Because the only possible solution to all three equations does not satisfy the third equation, the system has zero solutions.
11. Answer
13
First rewrite each equation in standard form.
For the first equation:
\(6x-2(y-5)=20\)
Distribute:
\(6x-2y+10=20\)
Subtract 10 from both sides:
\(6x-2y=10\)
For the second equation:
\(4x+3y=px-9\)
Move the \(px\) term to the left:
\((4-p)x+3y=-9\)
A system has no solution when the lines are parallel but not the same line. Therefore, the coefficients of \(x\) and \(y\) must be proportional.
So,
\(\frac{6}{4-p}=\frac{-2}{3}\)
Cross multiply:
\(18=-8+2p\)
\(26=2p\)
\(p=13\)
The constants are not proportional because
\(\frac{10}{-9}\ne\frac{6}{-9}\)
so the lines are distinct and parallel. Therefore, the system has no solution when \(p=13\).
12. Answer
6
First rewrite each equation in standard form.
For the first equation:
\(4x-3y+8=2x+y-6\)
Subtract \(2x\) from both sides and add \(3y\) to both sides:
\(2x-4y=-14\)
Divide by 2:
\(x-2y=-7\)
For the second equation:
\(5x+2y=px-10\)
Move all variable terms to one side:
\((5-p)x+2y=-10\)
A system has no solution when the lines are parallel but distinct. Therefore, the coefficients of \(x\) and \(y\) must be proportional, but the constants must not be proportional.
The first equation has coefficients \(1\) and \(-2\). The second equation has coefficients \(5-p\) and \(2\).
Set the ratios equal:
\( rac{1}{5-p}= rac{-2}{2}\)
\( rac{1}{5-p}=-1\)
\(5-p=-1\)
\(p=6\)
When \(p=6\), the second equation becomes \(-x+2y=-10\), or equivalently \(x-2y=10\). Since the first equation is \(x-2y=-7\), the lines are parallel but different, so the system has no solution.