Linear Equations, One Unknown - GMAT Quantitative

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Question

For what value of $\small N$ would the following equation have no solution?

$\small \small 3 (4x - 7) + 12 = 2 (5x - 3) + N (x - 3)$

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Answer

Simplify both sides of the equation as much as possible, and solve for $\small x$ in the equation in terms of $\small N$ :

$\small \small \small 3 (4x - 7) + 12 = 2 (5x - 3) + N (x - 3)$

$\small 3 \cdot 4x - 3 \cdot 7 + 12 = 2 \cdot 5x - 2 \cdot 3 + N \cdot x - N \cdot 3$

$\small 12x - 21 + 12 = 10x - 6 + Nx - 3N$

$\small 12x - 9 = (10+N)x + (- 6 - 3N)$

$\small \small 12x - (10+N)x = (- 6 - 3N) + 9$

$\small (2-N)x = 3 - 3N$

$\small x = \frac{3 - 3N}{2 -N}$

$\small x$ has exactly one solution unless the denominator is 0 - that is, $\small N = 2$ . We make sure that this value renders no solution by substituting:

$\small \small \small 3 (4x - 7) + 12 = 2 (5x - 3) + N (x - 3)$

$\small \small \small \small 3 (4x - 7) + 12 = 2 (5x - 3) + 2 (x - 3)$

$\small 12x - 21 + 12 = 10x - 6 + 2x - 6$

$\small 12x - 9 = 12x - 12$

$\small - 9 = - 12$

The equation has no solution, and $\small N = 2$ is the correct answer.

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Question

What is the midpoint coordinate of and ?

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Answer

Midpoint formula:

$\left(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2}\right)$

$(1,4)\ and\ (7,10)$

$\left(\frac{1+7}{2},\frac{4+10}{2}\right)$

$\left(\frac{8}{2},\frac{14}{2}\right)$

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Question

What is the midpoint coordinate of and ?

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Answer

Midpoint formula:

$\left(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2}\right)$

$(1,2)\ and\ (5,2)$

$\left(\frac{1+5}{2},\frac{2+2}{2}\right)$

$\left(\frac{6}{2},\frac{4}{2}\right)$

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Question

What is the midpoint coordinate of and ?

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Answer

Midpoint formula:

$\left(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2}\right)$

$(-2,-1)\ and\ (-8,7)$

$\left(\frac{-2+-8}{2},\frac{-1+7}{2}\right)$

$\left(\frac{-10}{2},\frac{6}{2}\right)$

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Question

Solve for :

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Answer

$z=\frac{1}{5}$

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Solve for :

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Solve for :

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Solve for :

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Question

Solve the following equation:

$2\left | x-5\right |+16=30$ .

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Answer

We start by isolating the absolute value expression:

$2\left | x-5\right |+16=30 \Leftrightarrow 2\left | x-5\right |=30-16=14\Leftrightarrow \left | x-5\right |=7$

This gives us two cases when we remove the absolute value:

and

Then we solve for each case:

$x - 5 = 7 \Rightarrow x = 7 + 5 \Rightarrow x= 12$

$x - 5 = -7 \Rightarrow x = -7 + 5\Rightarrow x= -2$

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Question

Solve for :

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Answer

$\left (5 N - 30 \right ) - \left (2 N + 8 \right )= 7N + 35$

$N = \frac{73}{-4}= - 18 \frac{1}{4}$

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Question

Solve for :

$\left (4x + 7 \right ) + 2 (x + 15 ) = 3(x-17)$

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Answer

$\left (4x + 7 \right ) + 2 (x + 15 ) = 3(x-17)$

$\left (4x + 7 \right ) + (2x+30) = 3x-51$

$x = \frac{-88}{3}= -29 \frac{1}{3}$

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Question

Which of the following equations has the set of all real numbers as its solution set?

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Answer

The right side of each equation is , which simplifies by way of distribution to

$8 (N+3) = 8 \cdot N + 8\cdot 3 = 8N + 24$

If the left side of the equation simplifies to an identical expression, the equation has all real numbers as its solutions.

We test the left side of each equation:

$2(N+4) + 6N = 2 \cdot N + 2 \cdot 4 + 6N = 2N + 8 + 6N = 8N + 8$

$3(N+4) + 5N= 3 \cdot N + 3 \cdot 4 + 5N = 3N + 12 + 5N = 8N +12$

\

$4(N+4) +4 N = 4 \cdot N + 4 \cdot 4 + 4N = 4N + 16 + 4N = 8N + 16$

$5(N+4) + 3N = 5 \cdot N + 5 \cdot 4 + 3N = 5N + 20 + 3N = 8N + 20$

$6(N+4) + 2N = 6 \cdot N + 6 \cdot 4 + 2N = 6N + 24 + 2N = 8N + 24$

Of the given choices,

can be rewritten as

,

which is an identity and has the set of all real numbers as its solution set.

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Question

Consider the incomplete equation

$10(N - 18) + 5N = \square (N-12 )$

What number replaces the box in order to form an equation with no solution?

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Answer

Set to be the number that replaces the box.

Simplify first:

Now solve for in terms of :

$\left (15 -A \right )N= 180 - 12A$

$N=\frac{ 180 - 12A}{15 -A }$

The only possible value of that might preclude the existence of a solution is , since it makes the denominator 0. However, let us test this value in the original equation:

As it turns out, replacing the box with 15 yields an identity, not a contradiction, so the solution set is the set of all real numbers. There is no number that fits the description.

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Question

Consider the incomplete equation

$4x + 17 = 6\left ( \square x - 16 \right ) - 4x$

Which of the following numbers can replace the box to form an equation whose one and only solution is 2?

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Answer

Rewrite this equation as

$4x + 17 = 6\left ( A x - 16 \right ) - 4x$

If 2 is a solution of the equation, then we can substitute 2 for to make a true arithmetic equation. Replace with 2 and solve for :

$4 \cdot 2 + 17 = 6\left ( A \cdot 2- 16 \right ) - 4 \cdot 2$

$8+ 17 = 6\left (2 A - 16 \right ) - 8$

$25 = 12 A - 96 \right ) - 8$

$A = \frac{129}{12}$

This number replaces the box in order to form the equation

$4x + 17 = 6\left ( \frac{129}{12}x - 16 \right ) - 4x$

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Question

Define a function as follows:

If , evaluate .

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Answer

Since , we can plug N in for x and 47 in for f(N) to get the following equation,

From here, we want to solve for N therefore we must isolate N on one side of the equation and all other numbers on the other side.

$N = 82 \div 8$

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Question

Solve the following equation for

$\frac{6x}{4} +7 = 18 +x$

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Answer

We proceed as follows

$\frac{6x}{4} +7 = 18 +x$ (Start)

(Multiply both sides by 4. Remember to distribute the 4 to both summands on both sides.)

(Subtract 28 from both sides)

(Subtract 4x from both sides)

(Divide both sides by 2)

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Question

Lisa went to a bargain bookstore where books were sold for dollars and magazines for dollars. After buying six books and four magazines, she only spent $30.00. How much did the books and magazines cost?

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Answer

We must first write out the equation to this problem.

If we set to be the cost of books then we can set to represent the cost of magazines.

She purchased books which means she spent on books.

She also bought magazines which means she spent on magazines.

The total spent was so when these two values are added together they must equal :

We can now solve for :

Remember: the cost of magazines was so we must plug in the value to find our answer.

So, our answer is that books cost $ and magazines cost $.

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Question

Students at a local college decide to make and sell t-shirts in order to raise money for their activities. They will pay the manufacturer $\$6.00$ per t-shirt made and a fixed fee of $\$200.00$ . If they sell each t-shirt for $\$16.00$ , how many t-shirts would they have to make and sell to raise $\$1600.00$ ?

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Answer

We set up the following equation:

The students will need to make and sell 180 t-shirts in order to raise $\$1600.00$ .

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Question

What is the value of in the following equation when ?

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Answer

When solving algebraic equations with one unknown, it is often easier to rearrange the equation first so that you have the unknown variable isolated.

So this:

becomes this when we subtract from both sides:

Then, divide both sides by to get all by itself:

$\frac{y-49}{-7}=x$

To finish, simply plug in for and simplify.

$x=\frac{-343-49}{-7}=\frac{-392}{-7}=56$

Thus, is our answer!

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Question

Solve for :

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Answer

In order to solve for , isolate on one side of the equation:

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