GMAT Math : Solving linear equations with two unknowns

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Example Questions

Example Question #1 : Algebra

Which equation is linear?

Possible Answers:

$\frac{y+8}{x-2}=x+6$

5x+7y-8yz=16

$y=x^{2}-55x+1$

$x+\Pi y+ez=log(5)$

none of them are linear

Correct answer:

$x+\Pi y+ez=log(5)$

Explanation:

Let's go through all of the answer choices.

1. $x+\Pi y+ez=log(5)$ : $\Pi$ and $e$ are both constants, so the equation is actually linear.

2. 5x + 7y - 8yz = 16: This is not linear because of the yz term.

3. $\frac{y+8}{x-2}=x+6$ : This can be transformed into y + 8 = (x + 6)(x - 2). Clearly when this is expanded, there will be an $x^{2}$ term, so this is not linear.

4. $y=x^{2}-55x+1$ : This is not linear either, also because of the $x^{2}$ term.

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Example Question #1 : Algebra

Solve.

2x+3y=26

3x-4y=5

Possible Answers:

x=4; y=6

x=3; y=1

x=10; y=2

x=7; y=4

Correct answer:

x=7; y=4

Explanation:

Solve for in the first equation:

2x=26-3y

$x=13-\frac{3}{2}y$

Substitute into the second equation:

$3(13-\frac{3}{2}y)-4y=5$

$39-\frac{9}{2}y-4y=-34$

$-\frac{9}{2}y-4y=-34$

$-\frac{17}{2}y=-34$

$y=(-34)(-\frac{2}{17})=4$

Solve for .

2x+3(4)=26

2x+12=26

2x=14

x=7

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Example Question #1082 : Gmat Quantitative Reasoning

4x+y=27

2x-3y=-11

What is $x^{2} - y?$

Possible Answers:

Correct answer:

Explanation:

Solve the first equation to get y=27-4x

Substitute that into the second equation and get

2x-3*(27-4x) = -11

Solve the equation to get x=5 , then substitute that into the first equation to get y=7 .

Plugging those two values into $x^{2} - y$ , gives

$5^{2}-7=18$

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Example Question #2 : Algebra

Solve the system of equations:

$\frac{x}{3}-\frac{y}{4} =1$

-4x+3y = 6

Possible Answers:

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

(3, 0)

(0,2)

(3,4)

The system has no solution.

Correct answer:

The system has no solution.

Explanation:

Multiply both sides of the first equation by 12:

$\frac{x}{3}-\frac{y}{4} =1$

$12 \cdot \left (\frac{x}{3}-\frac{y}{4} \right ) =12 \cdot 1$

$\frac{12}{1} \cdot \frac{x}{3}-\frac{12}{1} \cdot \frac{y}{4} =12$

$4x-\3y =12$

Now, add both sides of the two equations:

$4x-\3y =12$

$\underline{-4x+3y = 6}$

0=18

Since this is impossible, the system of equations is inconsistent and thus has no solution.

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Example Question #1 : Algebra

$x^{2} - y ^{2} = 125$

x + y = 25

Give the solution set for .

Possible Answers:

Insufficient information is given to answer the question.

x = 10

x = 15

$x = -10 \textrm{ or } x = 10$

$x = -15 \textrm{ or } x = 15$

Correct answer:

x = 15

Explanation:

$x^{2} - y ^{2} = 125$

The expression on the left factors as the difference of squares:

(x+y )(x-y ) = 125

Since x + y = 25 , we can substitute:

25(x-y ) = 125

$25(x-y ) \div 25 = 125\div 25$

x-y = 5

We now have a system of linear equations to solve:

x-y = 5

$\underline{x + y = 25}$

$2x \;\;\;\;\;=30$

$2x \div 2 =30\div 2$

x = 15

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Example Question #1 : Solving Linear Equations With Two Unknowns

A company wants to ship some widgets. If the weight of the box plus one widget is 6 pounds, and the weight of the box plus two widgets is 10 pounds, then what is the weight of the box and the weight of the widget? Put the answer in an ordered pair such that the ordered pair is (box weight, widget weight).

Possible Answers:

$\left ( 1,5 \right )$

$\left ( 2,4 \right )$

$\left ( 4,2 \right )$

$\left ( 2,3 \right )$

$\left ( 2,2 \right )$

Correct answer:

$\left ( 2,4 \right )$

Explanation:

Let the weight of the box be represented by and the weight of the widget be represented by . Since the weight of the box plus the weight of one widget is 6 pounds, this can be represented by the equation

B+W=6

Since the weight of the box plus two widgets is 10 pounds, this can be represented by the equation

B+2W=10

We now have two equations and two unknowns and we can now solve for and . To do this we solve the first equation for and substitute it into the second equation. Solving the first equation for we get

B=6-W

Substituting this into the second equation we get

$(6-W)+2W=10\Rightarrow 6-W+2W=10\Rightarrow$

$6+W=10\Rightarrow W=10-6=4.$

Using W=4 and substituting it into the first equation we get

$B+4=6\Rightarrow B=6-4=2.$ So the weight of the box is 2 pounds and the weight of the widget is 4 pounds. This gives us the ordered pair $\left ( 2,4 \right )$ .

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Example Question #5 : Algebra

Solve for when y=1.

7y+2x-8=9

Possible Answers:

Correct answer:

Explanation:

Plug in the given value and then isolate .

7y+2x-8=9

7(1)+2x-8=9

7+2x-8=9

2x-1=9

2x=9+1

2x=10

x=5

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Example Question #5 : Algebra

Whats the value of when x=20 :

6y+7=17+7x

Possible Answers:

Correct answer:

Explanation:

6y+7=17+7x

6y+7=17+7(20)

6y+7=17+140

6y=17+140-7

6y=150

y=25

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Example Question #4 : Algebra

Solve for :

2x+4y=10

Possible Answers:

x=-2y-5

x=2y-5

x=-2y+5

x=2y+5

Correct answer:

x=-2y+5

Explanation:

2x+4y=10

2x=10-4y

$x=\frac{10-4y}{2}$

x=-2y+5

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Example Question #7 : Solving Linear Equations With Two Unknowns

$x^{2}+ y^{2} = 20$

x+ 2y = 10

What is ?

Possible Answers:

y = 8

y = 5

y = 2

y = 4

y = 10

Correct answer:

y = 4

Explanation:

From the second equation:

x+ 2y = 10

x+ 2y - 2y = 10 - 2y

x = 10 - 2y

Substitute into the first, then solve:

$\left ( 10-2y\right )^{2}+ y^{2} = 20$

$10^{2} - 2 \cdot 10 \cdot 2y + \left ( 2y \right )^{2} + y^{2} = 20$

$4y ^{2} + y^{2} - 40y + 100 = 20$

$5 y^{2} - 40y + 100 = 20$

$5 y^{2} - 40y + 100 - 20= 20 - 20$

$5 y^{2} - 40y + 80= 0$

$5 \left ( y^{2} - 8y + 16 \right )= 0$

$5 \left ( y - 4 \right )^{2}= 0$

y - 4 = 0

y = 4

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GMAT Math : Solving linear equations with two unknowns

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #1 : Algebra

Example Question #1 : Algebra

Example Question #1082 : Gmat Quantitative Reasoning

Example Question #2 : Algebra

Example Question #1 : Algebra

Example Question #1 : Solving Linear Equations With Two Unknowns

Example Question #5 : Algebra

Example Question #5 : Algebra

Example Question #4 : Algebra

Example Question #7 : Solving Linear Equations With Two Unknowns

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