Solving linear equations with two unknowns - GMAT Quantitative Reasoning

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Question

Which equation is linear?

Answer

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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Question

Solve the system of equations.

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

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

Answer

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

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Question

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

Answer

Solve the first equation to get

Substitute that into the second equation and get

Solve the equation to get , then substitute that into the first equation to get .

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

$5^{2}-7=18$

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Question

Solve.

Answer

Solve for in the first equation:

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

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 .

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Question

Solve the system of equations:

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

Answer

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:

$4x-\3y =12$

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

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

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Question

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

Answer

Subtract the first equation from the second:

$x+y = 4\Rightarrow y=4-x$

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

$4x + 3(4-x) = 19\Rightarrow 4x+12-3x = 19\Rightarrow x+12 = 19$

Thus we get and .

Therefore is positive and is negative.

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Question

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

Give the solution set for .

Answer

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

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

Since , we can substitute:

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

We now have a system of linear equations to solve:

$\underline{x + y = 25}$

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

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Question

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).

Answer

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

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

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

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 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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Question

Solve for when

Answer

Plug in the given value and then isolate .

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Question

Whats the value of when :

Answer

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Question

Solve for :

Answer

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

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Question

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

What is ?

Answer

From the second equation:

Substitute into the first, then solve:

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

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

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

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

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

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

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

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

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Question

If and ; what is the value of ?

Answer

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,

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Question

Find the point of intersection of the two lines.

Answer

The correct answer is $\left(2,-\frac{1}{6}\right)$

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.

$y=-\frac{1}{6}$

Hence the point of intersection of the two lines is $\left(2,-\frac{1}{6}\right)$ .

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Question

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

Answer

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.

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Question

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

Answer

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:

$\small m=\frac{y-y'}{x-x'}=\frac{6-4}{14--6}=\frac{2}{20}=\frac{1}{10}$

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

Recall the general form for a linear equation:

$\small y=mx+b$

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

$\small \small y-mx=b \rightarrow b=y-mx$

$\small b=6-14*\frac{1}{10}{}=6-1.4=4.6$

So, we have our slope, $\frac{1}{10}$ and our -intercept, .

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Question

Solve the following system of linear equations:

Answer

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:

$-3y+x=2\rightarrow x=3y+2$

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

Now we can solve this equation for .

$-10y=2\rightarrow y=-\frac{1}{5}$

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

$x=3(-\frac{1}{5})+2=-\frac{3}{5}+2=-\frac{3}{5}+\frac{10}{5}=\frac{7}{5}$

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Question

Given and , find the values of and .

Answer

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:

___________________

We can now find by substituting our into any equation:

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Question

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

Answer

We have enough information to write out two equations:

$n\cdot p=24$

Using the first equation, we can narrow our potential values to: $\textup{ 1 and 24, 2 and 12, 3 and 8, and 4 and 6}$ .

Using the second equation, we can narrow down our values even further to $\textup{3 and 8}$ . 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.

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Question

Solve for .

Answer

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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