GMAT Math : Algebra

Study concepts, example questions & explanations for GMAT Math

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Example Questions

Example Question #11 : Algebra

Solve the system of equations.

\dpi{100} \small 4x+3y=7

\dpi{100} \small 7x-14y=21

Possible Answers:

\dpi{100} \small x=\frac{23}{11}\and\ y=\frac{5}{11}

\dpi{100} \small x=\frac{-23}{11}\and\ y=\frac{5}{11}

\dpi{100} \small x=\frac{23}{11}\and\ y=\frac{-5}{11}

x = all real numbers,

y = all real numbers

\dpi{100} \small x=\frac{-23}{11}\and\ y=\frac{-5}{11}

Correct answer:

\dpi{100} \small x=\frac{23}{11}\and\ y=\frac{-5}{11}

Explanation:

Let's first look at the 2nd equation. All three terms in \dpi{100} \small 7x-14y=21 can be divided by 7.  Then \dpi{100} \small x-2y=3  We can isolate x to get \dpi{100} \small x=3+2y

Now let's plug \dpi{100} \small x=3+2y into the 1st equation, \dpi{100} \small 4x+3y=7:

\dpi{100} \small 4\left ( 3+2y \right )+3y=7

\dpi{100} \small 12+8y+3y=7

\dpi{100} \small 11y=-5

\dpi{100} \small y=\frac{-5}{11}  Now let's plug our y-value into \dpi{100} \small x=3+2y to solve for y:

\dpi{100} \small x=3+2\left\left ( \frac{-5}{11} \right )=3-\frac{10}{11}=\frac{33}{11}-\frac{10}{11}=\frac{23}{11}

So \dpi{100} \small x=\frac{23}{11}\and\ y=\frac{-5}{11}

Example Question #12 : Algebra

Choose the statement that most accurately describes the system of equations.

Possible Answers:

is positive, is positive.

is negative, is negative.

is negative, is positive.

No unique solution.

is positive, is negative.

Correct answer:

is positive, is negative.

Explanation:

Subtract the first equation from the second:

Now we can substitute this into either equation. We'll plug it into the first equation here:

Thus we get and .

Therefore is positive and is negative.

Example Question #13 : Solving Linear Equations With Two Unknowns

If  and ; what is the value of ?

Possible Answers:

Correct answer:

Explanation:

For this problem we can use the elimination method to solve for one of our variables. We do this my multiplying our first equation by -2.

From here we can combine this equation with our second equation given in the question and solve for x.

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Now we plug 1 back into our original equation and solve for y.

Therefore,

 

Example Question #14 : Algebra

Find the point of intersection of the two lines.

Possible Answers:

None of the other answers

Correct answer:

None of the other answers

Explanation:

The correct answer is 

 

There are a few ways of solving this. The method I will use is the method of elimination.

(Start)

 

(Multiply the 2nd equation by -1 and add the result to the first equation, combining like terms. Now the top equation simplifies to

 

Now that we have one of the variables solved for, we can plug into either of the original equations, and we can get our , Let's use the 2nd equation.

 

 

Hence the point of intersection of the two lines is .

 

Example Question #11 : Algebra

Julie has  coins, all dimes and quarters. The total value of all her coins is . How many dimes and quarters does Julie have?

Possible Answers:

 quarters and  dimes

 quarters and  dimes

 quarters and  dimes

 quarters and  dimes

 quarters and  dimes

Correct answer:

 quarters and  dimes

Explanation:

Let  be the number of dimes Julie has and  be the numbers of quarters she has. The number of dimes and the number of quarters add up to  coins. The value of all quarters and dimes is . We can then write the following system of equations:

To use substitution to solve the problem, begin by rearranging the first equation so that  is by itself on one side of the equals sign:

Then, we can replace  in the second equation with :

Distribute the :

Subtract  from each side of the equation:

Divide each side of the equation by :

Now, we can insert our value for  into the first equation and solve for :

Julie has  quarters and  dimes.

Example Question #16 : Algebra

Solve the following system of linear equations:

Possible Answers:

Correct answer:

Explanation:

To solve a system of two equations with two unknowns, we first solve one of the equations for one of the variables and then substitute that value into the other equation. This allows us to find a solution for one of the variables, which we then plug back into either equation to find the solution for the other variable:

Substituting the right side of the rearranged equation into the other equation for , we get:

Now we can solve this equation for .

Now that we know the value of , we can plug that value into the other equation for  and solve for :

Example Question #1094 : Gmat Quantitative Reasoning

 is a linear equation that passes through the points  and . What is the slope , and y-intercept  of ?

Possible Answers:

Correct answer:

Explanation:

We're told to find the slope and -intercept of a line that passes through the points  and . To begin, calculate the slope using the following equation:

So now that we have our slope, we need to find our -intercept.

Recall the general form for a linear equation:

Rearrange to solve for  and use our slope and one of the given points to solve:

So, we have our slope,  and our -intercept, .

Example Question #17 : Algebra

Given  and , find the values of  and .

Possible Answers:

Correct answer:

Explanation:

We can solve this problem by setting up a system of equations and using elimination:

We can eliminate the  and solve for  by multiplying the bottom equation by  and adding the equations:

   

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We can now find  by substituting our  into any equation:

Example Question #1101 : Gmat Quantitative Reasoning

The product of two positive numbers,  and , yields . If their sum is , what is the value of 

Possible Answers:

Correct answer:

Explanation:

We have enough information to write out two equations:

Using the first equation, we can narrow our potential values to:.

Using the second equation, we can narrow down our values even further to  We are, however, being asked specifically for the value of . Since we cannot state if the  or the  represents  and which represents , we cannot answer this question. Additional data, such as  is less than , would be required.

Example Question #14 : Algebra

Solve for .

Possible Answers:

Correct answer:

Explanation:

We can solve this problem in the same way we would solve a system of equations using elimination. Since we are solving for  we can manipulate the system to cancel out the  values:                                             

                                 

We then add the equations. Notice how the  values cancel out 

leaving us with 

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