College Algebra - Systems of Equations

Solve the following system of equations:

$y=x^{2}+1$

Explanation

$y=x^{2}+1$

Subtract the second equation from the first, to eliminate the y terms:

$y=x^{2}+1$

This yields the following: $0=x^{2}-2x$

Factor out an x from both terms:

Solve for x:

Plug in the values for x into the first equation:

$y=(0)^{2}+1=1$

$y=(2)^{2}+1=4+1=5$

Therefore, when , , and when ,

Plug in the values for x and y to make sure they satisfy the second equation too:

Solution:

Solve the system of equations.

None of the other answers are correct.

Explanation

Isolate in the first equation.

Plug into the second equation to solve for .

Plug into the first equation to solve for .

Now we have both the and values and can express them as a point: .

Solve the system of equations.

None of the other answers are correct.

Explanation

Isolate in the first equation.

Plug into the second equation to solve for .

Plug into the first equation to solve for .

Now we have both the and values and can express them as a point: .

Solve the following system of equations:

$x=2y^{2}$

$(2,-1)(\frac{1}{2},\frac{1}{2})$

$(2,1)(\frac{1}{2},\frac{1}{2})$

$(-2,-1)(\frac{1}{2},\frac{1}{2})$

$(2,-1)(\frac{-1}{2},\frac{1}{2})$

Explanation

$x=2y^{2}$

Substitute the value of y from the second equation, into the first equation:

$x=2y^{2}=2(-x+1)^{2}=2(x^{2}-2x+1)=2x^{2}-4x+2$

Subtract x from both sides of the equation:

$2x^{2}-5x+2=0$

Use the quadratic formula to solve for x: $x=\frac{5\pm \sqrt{25-4(2)(2)}}{4}=\frac{5\pm \sqrt{25-16}}{4}=\frac{5\pm \sqrt{9}}{4}=\frac{5\pm 3}{4}$

or $x=\frac{1}{2}$

Plug the first value of x into the second equation:

Plug the second value of x into the second equation: $y=-(\frac{1}{2})+1=\frac{1}{2}$

or $x=\frac{1}{2},y=\frac{1}{2}$

Verify the first solution by plugging the first values of x and y into both equations:

$(2)=2(-1)^{2}=2$

Verify the second solution by plugging the second values of x and y into both equations:

$(\frac{1}{2})=2(\frac{1}{2})^{2}=2(\frac{1}{4})=\frac{1}{2}$

$(\frac{1}{2})=-(\frac{1}{2})+1=\frac{1}{2}$

Solution:

$(2,-1)(\frac{1}{2},\frac{1}{2})$

Solve the following system of equations:

$y=x^{2}+1$

Explanation

$y=x^{2}+1$

Subtract the second equation from the first, to eliminate the y terms:

$y=x^{2}+1$

This yields the following: $0=x^{2}-2x$

Factor out an x from both terms:

Solve for x:

Plug in the values for x into the first equation:

$y=(0)^{2}+1=1$

$y=(2)^{2}+1=4+1=5$

Therefore, when , , and when ,

Plug in the values for x and y to make sure they satisfy the second equation too:

Solution:

Solve the following system of equations:

$x=2y^{2}$

$(2,-1)(\frac{1}{2},\frac{1}{2})$

$(2,1)(\frac{1}{2},\frac{1}{2})$

$(-2,-1)(\frac{1}{2},\frac{1}{2})$

$(2,-1)(\frac{-1}{2},\frac{1}{2})$

Explanation

$x=2y^{2}$

Substitute the value of y from the second equation, into the first equation:

$x=2y^{2}=2(-x+1)^{2}=2(x^{2}-2x+1)=2x^{2}-4x+2$

Subtract x from both sides of the equation:

$2x^{2}-5x+2=0$

Use the quadratic formula to solve for x: $x=\frac{5\pm \sqrt{25-4(2)(2)}}{4}=\frac{5\pm \sqrt{25-16}}{4}=\frac{5\pm \sqrt{9}}{4}=\frac{5\pm 3}{4}$

or $x=\frac{1}{2}$

Plug the first value of x into the second equation:

Plug the second value of x into the second equation: $y=-(\frac{1}{2})+1=\frac{1}{2}$

or $x=\frac{1}{2},y=\frac{1}{2}$

Verify the first solution by plugging the first values of x and y into both equations:

$(2)=2(-1)^{2}=2$

Verify the second solution by plugging the second values of x and y into both equations:

$(\frac{1}{2})=2(\frac{1}{2})^{2}=2(\frac{1}{4})=\frac{1}{2}$

$(\frac{1}{2})=-(\frac{1}{2})+1=\frac{1}{2}$

Solution:

$(2,-1)(\frac{1}{2},\frac{1}{2})$

A man in a canoe travels upstream 400 meters in 2 hours. In the same canoe, that man travels downstream 600 meters in 2 hours.

What is the speed of the current, , and what is the speed of the boat in still water, ?

$c=50\frac{m}{hr};\ b=250\frac{m}{hr}$

$c=250\frac{m}{hr};\ b=50\frac{m}{hr}$

$c=100\frac{m}{hr};\ b=200\frac{m}{hr}$

$c=75\frac{m}{hr};\ b=225\frac{m}{hr}$

More information is needed

Explanation

This problem is a system of equations, and uses the equation $\text{distance}=\text{rate}\times\text{time}$ .

Start by assigning variables. Let stand for the rate of the boat, let stand for the rate of the current.

When the boat is going upstream, the total rate is equal to . You must subtract because the rates are working against each other—the boat is going slower than it would because it has to work against the current.

Using our upstream distance (400m) and time (2hr) from the question, we can set up our rate equation:

$400 = (b-c) \times 2$

When the boat is going downstream, the total rate is equal to because the boat and current are working with each other, causing the boat to travel faster.

We can refer to the downstream distance (600m) and time (2hr) to set up the second equation:

$600 = (b+c) \cdot 2$

From here, use elimination to solve for and .

1. Set up the system of equations, and solve for .

2. Subsitute into one of the equations to solve for .

Solve for : $\begin{bmatrix} 2x+5y =1\ 3x+5y =4 \end{bmatrix}$

Explanation

We can evaluate the value of by subtracting the first equation from the second equation since both equations share , and can be eliminated.

The equation becomes:

Substitute this value back into either the first or second equation, and solve for y.

The answer is:

A man in a canoe travels upstream 400 meters in 2 hours. In the same canoe, that man travels downstream 600 meters in 2 hours.

What is the speed of the current, , and what is the speed of the boat in still water, ?

$c=50\frac{m}{hr};\ b=250\frac{m}{hr}$

$c=250\frac{m}{hr};\ b=50\frac{m}{hr}$

$c=100\frac{m}{hr};\ b=200\frac{m}{hr}$

$c=75\frac{m}{hr};\ b=225\frac{m}{hr}$

More information is needed

Explanation

This problem is a system of equations, and uses the equation $\text{distance}=\text{rate}\times\text{time}$ .

Start by assigning variables. Let stand for the rate of the boat, let stand for the rate of the current.

Using our upstream distance (400m) and time (2hr) from the question, we can set up our rate equation:

$400 = (b-c) \times 2$

When the boat is going downstream, the total rate is equal to because the boat and current are working with each other, causing the boat to travel faster.

We can refer to the downstream distance (600m) and time (2hr) to set up the second equation:

$600 = (b+c) \cdot 2$

From here, use elimination to solve for and .

1. Set up the system of equations, and solve for .

2. Subsitute into one of the equations to solve for .

What is a solution to this system of equations?

$\left{\begin{matrix} 2y=3x+11\ y=-5x-14\end{matrix}\right.$