Systems of Equations - SAT Subject Test in Math II

Card 0 of 7

Question

Where do the graphs of these two equations intersect?

Equation 1:

Equation 2:

Answer

One can find the points of intersection of these two functions by setting them equal to one another, essentially substituting in one equation for the -side of the other equation. This will tell us when the (output) will be the same in each equation for a given (input).

By simplifying this equation and setting it equal to zero, we can find the two -values that produce the same values in the system of the two equations.

Subtract from both sides of the equation, and add to both sides.

Factoring this last equation makes it easier to find the -values that will result in zero on the left side of the equation. Set the two parenthetical phrases equal to zero to find two separate -values that satisfy the equation. These values will be the values of the points of intersection between the two equations.

We know our factors multiply to , and the six times one factor plus the other is equal to

and , so and are our factors.

$6x+1 = 0 \ \textup{and}\ x-1 = 0$

$x = -\frac{1}{6} \ \textup{and}\ x=1$

Substituting these two values into either of the two original equations results in the -values of the points of intersection.

$y_2=5(-\frac{1}{6})-2=-\frac{17}{6}$

$(1,3), (-\frac{1}{6}, -\frac{17}{6})$ are the points of intersection.

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Question

Find the solution:

Answer

To solve this system of equations, we must first eliminate one of the variables. We will begin by eliminating the variables by finding the least common multiple of the variable's coefficients. The least common multiple of 3 and 2 is 6, so we will multiply each equation in the system by the corresponding number, like

.

By using the distributive property, we will end up with

Now, add down each column so that you have

$0x+13y=39 \rightarrow 13y=39$

Then you solve for and determine that .

But you're not done yet! To find , you have to plug your answer for back into one of the original equations:

Solve, and you will find that .

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Question

$\frac{4}{a} + \frac{7}{b} = 18$

$\frac{3}{a} - \frac{5}{b} = 34$

Evaluate

Answer

Rewrite the two equations by setting $x = \frac{1}{a}$ and $y = \frac{1}{b}$ and substituting:

$\frac{4}{a} + \frac{7}{b} = 18$

$4 \cdot \frac{1}{a} + 7 \cdot \frac{1 }{b} = 18$

$\frac{3}{a} - \frac{5}{b} = 34$

$3 \cdot \frac{1}{a} - 5 \cdot \frac{1}{b} = 34$

The result is a two-by-two linear system in terms of and :

This can be solved, among other ways, using Gaussian elimination on an augmented matrix:

$\begin{bmatrix} 4& 7\ 3& -5 \end{matrix} \begin{vmatrix} 18 \ 34 \ \end{bmatrix}$

Perform row operations until the left two columns show identity matrix $\begin{bmatrix} 1 & 0\ 0 &1 \end{bmatrix}$ . One possible sequence:

$R1-R2 \rightarrow R1$

$\begin{bmatrix} 4 - 3 & 7- (-5 )\ 3& -5 \end{matrix} \begin{vmatrix} 18 - 34 \ 34 \ \end{bmatrix}$

$\begin{bmatrix} 1 & 12 \ 3& -5 \end{matrix} \begin{vmatrix} -16 \ 34 \ \end{bmatrix}$

$R2 - 3R1 \rightarrow R2$

$\begin{bmatrix} 1 & 12 \ 3 - 3(1)& -5 - 3(12) \end{matrix} \begin{vmatrix} -16 \ 34 - 3(-16) \ \end{bmatrix}$

$\begin{bmatrix} 1 & 12 \ 0& -41\end{matrix} \begin{vmatrix} -16 \ 82\ \end{bmatrix}$

$R2 \div (-41 )\rightarrow R2$

$\begin{bmatrix} 1 & 12 \ 0\div (-41 )& -41\div (-41 )\end{matrix} \begin{vmatrix} -16 \ 82\div (-41 )\ \end{bmatrix}$

$\begin{bmatrix} 1 & 12 \ 0& 1 \end{matrix} \begin{vmatrix} -16 \-2\ \end{bmatrix}$

$R1 - 12 R2 \rightarrow R1$

$\begin{bmatrix} 1 - 12(0) & 12 - 12(1) \ 0& 1 \end{matrix} \begin{vmatrix} -16 - 12(-2) \ 2\ \end{bmatrix}$

$\begin{bmatrix} 1 & 0 \ 0& 1 \end{matrix} \begin{vmatrix} 8 \ 2\ \end{bmatrix}$

and .

In the former equation, substitute $\frac{1}{a}$ back for :

$\frac{1}{a} = 8$ .

Taking the reciprocal of both sides, we get

$a = \frac{1}{8}$ .

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Question

$4 \sqrt [3] {a} + 7 \sqrt [3]{b} = 18$

$3 \sqrt [3]{a} -5 \sqrt [3]{b} = 34$

Evaluate .

Answer

Rewrite the two equations by setting $x = \sqrt [3] {a}$ and $y = \sqrt [3] {b}$ and substituting:

$4 \sqrt [3] {a} + 7 \sqrt [3]{b} = 18$

$3 \sqrt [3]{a} -5 \sqrt [3]{b} = 34$

The result is a two-by-two linear system in terms of and :

This can be solved, among other ways, using Gaussian elimination on an augmented matrix:

$\begin{bmatrix} 4& 7\ 3& -5 \end{matrix} \begin{vmatrix} 18 \ 34 \ \end{bmatrix}$

Perform row operations until the left two columns show identity matrix $\begin{bmatrix} 1 & 0\ 0 &1 \end{bmatrix}$ . One possible sequence:

$R1-R2 \rightarrow R1$

$\begin{bmatrix} 4 - 3 & 7- (-5 )\ 3& -5 \end{matrix} \begin{vmatrix} 18 - 34 \ 34 \ \end{bmatrix}$

$\begin{bmatrix} 1 & 12 \ 3& -5 \end{matrix} \begin{vmatrix} -16 \ 34 \ \end{bmatrix}$

$R2 - 3R1 \rightarrow R2$

$\begin{bmatrix} 1 & 12 \ 3 - 3(1)& -5 - 3(12) \end{matrix} \begin{vmatrix} -16 \ 34 - 3(-16) \ \end{bmatrix}$

$\begin{bmatrix} 1 & 12 \ 0& -41\end{matrix} \begin{vmatrix} -16 \ 82\ \end{bmatrix}$

$R2 \div (-41 )\rightarrow R2$

$\begin{bmatrix} 1 & 12 \ 0\div (-41 )& -41\div (-41 )\end{matrix} \begin{vmatrix} -16 \ 82\div (-41 )\ \end{bmatrix}$

$\begin{bmatrix} 1 & 12 \ 0& 1 \end{matrix} \begin{vmatrix} -16 \-2\ \end{bmatrix}$

$R1 - 12 R2 \rightarrow R1$

$\begin{bmatrix} 1 - 12(0) & 12 - 12(1) \ 0& 1 \end{matrix} \begin{vmatrix} -16 - 12(-2) \ 2\ \end{bmatrix}$

$\begin{bmatrix} 1 & 0 \ 0& 1 \end{matrix} \begin{vmatrix} 8 \ 2\ \end{bmatrix}$

and .

Substituting back in the second equation and solving for by cubing both sides:

$\sqrt [3] {b}= 2$

$(\sqrt [3] {b})^{3}= 2^{3}$

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Question

$4 \sqrt{ a} + 3 \sqrt{b} = 27$

$5 \sqrt{ a} + 2 \sqrt{b} = 25$

Evaluate .

Answer

Rewrite the two equations by setting $x = \sqrt {a}$ and $y = \sqrt {b}$ and substituting:

$4 \sqrt{ a} + 3 \sqrt{b} = 27$

$5 \sqrt{ a} + 2 \sqrt{b} = 25$

In terms of and , this is a two-by-two linear system:

Rewrite this as an augmented matrix as follows:

$\begin{bmatrix} 4& 3\ 5& 2 \end{matrix} \begin{vmatrix} 27\ 25 \end{bmatrix}$

Perform the following row operations to make the left two columns the identity matrix $\begin{bmatrix}1 & 0\ 0 &1 \end{bmatrix}$ :

$R1 \leftrightarrow R2$

$\begin{bmatrix} 5& 2 \ 4& 3 \end{matrix} \begin{vmatrix} 25 \ 27 \end{bmatrix}$

$R1-R2 \rightarrow R1$

$\begin{bmatrix} 5 -4& 2 -3 \ 4& 3 \end{matrix} \begin{vmatrix} 25 -27 \ 27 \end{bmatrix}$

$\begin{bmatrix} 1& -1 \ 4& 3 \end{matrix} \begin{vmatrix} -2 \ 27 \end{bmatrix}$

$R2 - 4R1 \rightarrow R2$

$\begin{bmatrix} 1& -1 \ 4 - 4(1)& 3- 4(-1) \end{matrix} \begin{vmatrix} -2 \ 27-4(-2) \end{bmatrix}$

$\begin{bmatrix} 1& -1 \ 0& 7 \end{matrix} \begin{vmatrix} -2 \ 35 \end{bmatrix}$

$R2 \div 7 \rightarrow R2$

$\begin{bmatrix} 1& -1 \ 0 \div 7 & 7 \div 7 \end{matrix} \begin{vmatrix} -2 \ 35 \div 7 \end{bmatrix}$

$\begin{bmatrix} 1& -1 \ 0 & 1 \end{matrix} \begin{vmatrix} -2 \ 5 \end{bmatrix}$

$R1 +R2 \rightarrow R1$

$\begin{bmatrix} 1+ 0& -1+1 \ 0 & 1 \end{matrix} \begin{vmatrix} -2+5 \ 5 \end{bmatrix}$

$\begin{bmatrix} 1 & 0 \ 0 & 1 \end{matrix} \begin{vmatrix} 3 \ 5 \end{bmatrix}$

and .

Replacing with $\sqrt {a}$ , then squaring both sides:

$\sqrt {a} = 3$

$(\sqrt {a}) ^{2}= 3^{2}$

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Question

Solve for : $\begin{bmatrix} 2x-8y=3\x+5y = 1 \end{bmatrix}$

Answer

To solve for , we will need to eliminate the variable. To do so, we can choose either the substitution or the elimination method.

Let's choose the substitution method. Rewrite the second equation so that is isolated.

Subtract on both sides.

Divide by five on both sides.

$\frac{5y}{5} = \frac{-x+1}{5}$

$y=- \frac{1}{5}x+\frac{1}{5}$

Substitute this back to the first equation.

$2x-8(- \frac{1}{5}x+\frac{1}{5})=3$

Distribute the terms.

$2x+\frac{8}{5}x-\frac{8}{5} = 3$

Multiply both sides by 5 to eliminate the fractions.

$5(2x+\frac{8}{5}x-\frac{8}{5}) = 3\cdot 5$

Add 8 on both sides and combine like-terms on the left.

Divide by 18 on both sides.

$\frac{18x}{18} = \frac{23}{18}$

The answer is: $\frac{23}{18}$

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Question

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

Answer

In order to solve for the x-variable, we will need to use the elimination method to eliminate the y-variable.

Multiply the second equation by two.

$\begin{bmatrix} 3x+2y = 1\ x-y=-1 \end{bmatrix} \rightarrow\begin{bmatrix} 3x+2y = 1\ 2x-2y=-2 \end{bmatrix}$

Add both equations.

Divide by five on both sides.

$\frac{5x}{5}=\frac{-1}{5}$

The answer is: $-\frac{1}{5}$

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