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ISEE Upper Level Math : How to find the solution to an equation

Study concepts, example questions & explanations for ISEE Upper Level Math

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All ISEE Upper Level Math Resources

6 Diagnostic Tests 244 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #61 : How To Find The Solution To An Equation

Solve the set of equations:

3x+y=5

-4x+2y=-10

Possible Answers:

x=2, y=-1

x=-2,y=1

x=-1,y=1

x=-2,y=-1

x=2,y=1

Correct answer:

x=2, y=-1

Explanation:

Solve the first equation for :

$3x+y=5\Rightarrow y=5-3x$

Substitute into the second equation:

$-4x+2y=-10\Rightarrow -4x+2(5-3x)=-10\Rightarrow -4x+10-6x=-10$

$\Rightarrow -10x=-20\Rightarrow x=2$

PLug this back into our new equation for :

$y=5-3x\Rightarrow y=5-3\times 2=5-6\Rightarrow y=-1$

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Example Question #62 : How To Find The Solution To An Equation

Solve the set of equations:

7x+y=50

-2x+5y=-9

Possible Answers:

x=-7,y=-1

x=7,y=7

x=-7,y=1

x=7,y=-1

x=7,y=1

Correct answer:

x=7,y=1

Explanation:

Solve the first equation for :

$7x+y=50\Rightarrow y=50-7x$

Substitute into the second equation:

$-2x+5(50-7x)=-9\Rightarrow -2x+250-35x=-9$

$\Rightarrow -37x=-259\Rightarrow x=\frac{-259}{-37}=7$

Plug this back into our new equation for :

$y=50-7x\Rightarrow y=50-7\times 7=50-49=1$

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Example Question #63 : How To Find The Solution To An Equation

Solve the set of equations:

2a+b=4

b+2c=10

a+c=5

Possible Answers:

a=1,b=2,c=4

a=1,b=4,c=2

a=1,b=2,c=2

a=1,b=4,c=4

a=2,b=4,c=4

Correct answer:

a=1,b=2,c=4

Explanation:

Solve the first equation for :

$2a+b=4\Rightarrow b=4-2a$

Substitute this expression into the second equation:

$b+2c=10\Rightarrow (4-2a)+2c=10\Rightarrow -2a+2c=6$

Now we should solve the following set of equations for and :

-2a+2c=6

a+c=5

Solve the second equation for :

$a+c=5\Rightarrow a=5-c$

Substitute into the first equation:

$-2a+2c=6\Rightarrow -2(5-c)+2c=6\Rightarrow -10+2c+2c=6\Rightarrow 4c=16$

$\Rightarrow c=4$

$a=5-c\Rightarrow a=5-4=1$

We also know b=4-2a , so plug in our new values:

$b=4-2a=4-2\times 1=2$

a=1, b=2, c=4

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Example Question #64 : How To Find The Solution To An Equation

Solve the set of equations:

-a+b=10

3b+c=43

4a-c=1

Possible Answers:

a=2,b=10,c=8

a=2,b=12,c=7

a=2, b=12, c=6

a=3,b=12,c=7

a=2,b=12,c=8

Correct answer:

a=2,b=12,c=7

Explanation:

Solve the first equation for :

$-a+b=10\Rightarrow b=10+a$

Substitute this expression into the second equation:

$3b+c=43\Rightarrow 3(10+a)+c=43\Rightarrow 30+3a+c=43\Rightarrow 3a+c=13$

Now we should solve the following set of equations:

4a-c=1

3a+c=13

We can sum up both equations to get:

$(4a-c)+(3a+c)=1+13\Rightarrow 7a=14\Rightarrow a=2$

Substitute this new value into the second equation of the second set i.e. 3a+c=13 :

$3a+c=13\Rightarrow 3\times 2+c=13\Rightarrow c=13-6=7$

We had b=10+a , so now:

b=10+a=10+2=12

a=2, b=12, c=7

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Example Question #61 : How To Find The Solution To An Equation

Solve for .

2x+7x-25=35-6x

Possible Answers:

3.5

4.5

Correct answer:

Explanation:

First we need to rewrite the problem by combining the like terms:

$2x+7x-25=35-6x\Rightarrow 9x-25=35-6x$

Now add to both sides:

$(9x+6x)-25=35-6x+6x\Rightarrow 15x-25=35\Rightarrow 15x=35+25$

$\Rightarrow 15x=60\Rightarrow x=\frac{60}{15}=4$

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Example Question #62 : How To Find The Solution To An Equation

Solve for .

9x^2-4=4x^2+76

Possible Answers:

$x=\pm 4$

x=5

$x=\pm 5$

x=3

$x=\pm 3$

Correct answer:

$x=\pm 4$

Explanation:

First we should combine like terms.

$9x^2-4=4x^2+76\Rightarrow (9x^2-4x^2)-4=(4x^2-4x^2)+76$

$\Rightarrow 5x^2-4=76$

$5x^2-4+4=76+4\Rightarrow 5x^2=80\Rightarrow x^2=\frac{80}{5}=16$

$x=\pm 4$

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Example Question #63 : How To Find The Solution To An Equation

Solve for .

$\frac{x}{8}=\frac{3}{4}-\frac{5}{6}$

Possible Answers:

$-\frac{3}{4}$

$\frac{3}{4}$

$\frac{2}{3}$

$-\frac{2}{3}$

$-\frac{1}{4}$

Correct answer:

$-\frac{2}{3}$

Explanation:

First we need to rewrite the right side fractions as equivalent fractions with the same denominators. The easiest denominator to use is the product of the denominators ( $4\times 6=24$ ). However, the best denominator to use is the least common denominator, which is the least common multiple of the denominators ( in this case).

$\frac{x}{8}=\frac{3}{4}-\frac{5}{6}\Rightarrow \frac{x}{8}=\frac{3}{4}*\frac{3}{3}-\frac{5}{6}*\frac{2}{2}$

$\Rightarrow \frac{x}{8}=\frac{9-10}{12}\Rightarrow \frac{x}{8}=-\frac{1}{12}\Rightarrow x=-\frac{8}{12}=-\frac{2}{3}$

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Example Question #2 : Word Problems

John sells cloths. His monthly salary is 1200 dollars plus a commission of dollars for each cloth sold. If he sells cloths this month, how much money does he make in dollars?

Possible Answers:

1800

1760

1700

1600

1750

Correct answer:

1760

Explanation:

In order to find how much he makes from selling the cloths, we first multiply the number of cloths sold by dollars.

$80\times 7=560$ dollars

Now we add his monthly salary:

560+1200=1760 dollars

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Example Question #2 : Word Problems

Tom works in a mall and sells shoes. His monthly salary is 1100 dollars plus a commission of dollars for every pair of shoes that he sells. How many pairs of shoes must he sell to earn 2060 dollars per month.

Possible Answers:

100

110

115

120

125

Correct answer:

120

Explanation:

Let number of pairs of shoes.

Then we can set up the following equation:

2060=1100+8x

Now we solve the equation for :

$\Rightarrow 2060-1100=(1100-1100)+8x\Rightarrow 960=8x\Rightarrow x=\frac{960}{8}=120$

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Example Question #26 : Percentages

A college has 1200 students. If percent of them are boys, how many girls are there?

Possible Answers:

840

300

800

360

350

Correct answer:

360

Explanation:

percent of the students are boys, so percent of the students are girls.

Let = number of girls.

Then we can set up the following proportion:

$\frac{30}{100}=\frac{x}{1200}\Rightarrow \frac{3}{10}=\frac{x}{1200}\Rightarrow 10x=3\times 1200\Rightarrow x=\frac{3600}{10}=360$

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All ISEE Upper Level Math Resources

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ISEE Upper Level Math : How to find the solution to an equation

Study concepts, example questions & explanations for ISEE Upper Level Math

All ISEE Upper Level Math Resources

Example Questions

Example Question #61 : How To Find The Solution To An Equation

Example Question #62 : How To Find The Solution To An Equation

Example Question #63 : How To Find The Solution To An Equation

Example Question #64 : How To Find The Solution To An Equation

Example Question #61 : How To Find The Solution To An Equation

Example Question #62 : How To Find The Solution To An Equation

Example Question #63 : How To Find The Solution To An Equation

Example Question #2 : Word Problems

Example Question #2 : Word Problems

Example Question #26 : Percentages

All ISEE Upper Level Math Resources

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