# 5.2 Solve systems of equations by substitution  (Page 2/5)

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Solve the system by substitution. $\left\{\begin{array}{c}3x+y=5\hfill \\ 2x+4y=-10\hfill \end{array}$

## Solution

We need to solve one equation for one variable. Then we will substitute that expression into the other equation.

 Solve for y . Substitute into the other equation. Replace the y with −3 x + 5. Solve the resulting equation for x . Substitute x = 3 into 3 x + y = 5 to find y . The ordered pair is (3, −4). Check the ordered pair in both equations: $\begin{array}{cccc}\begin{array}{ccc}\hfill 3x+y& =\hfill & 5\hfill \\ \hfill 3·3+\left(-4\right)& \stackrel{?}{=}\hfill & 5\hfill \\ \hfill 9-4& \stackrel{?}{=}\hfill & 5\hfill \\ \hfill 5& =\hfill & 5\phantom{\rule{0.2em}{0ex}}✓\hfill \end{array}& & & \begin{array}{ccc}\hfill 2x+4y& =\hfill & -10\hfill \\ \hfill 2·3+4\left(-4\right)& =\hfill & -10\hfill \\ \hfill 6-16& \stackrel{?}{=}\hfill & -10\hfill \\ \hfill -10& =\hfill & -10\phantom{\rule{0.2em}{0ex}}✓\hfill \end{array}\end{array}$ The solution is (3, −4).

Solve the system by substitution. $\left\{\begin{array}{c}4x+y=2\hfill \\ 3x+2y=-1\hfill \end{array}$

$\left(1,-2\right)$

Solve the system by substitution. $\left\{\begin{array}{c}-x+y=4\hfill \\ 4x-y=2\hfill \end{array}$

$\left(2,6\right)$

In [link] it was easiest to solve for y in the first equation because it had a coefficient of 1. In [link] it will be easier to solve for x .

Solve the system by substitution. $\left\{\begin{array}{c}x-2y=-2\hfill \\ 3x+2y=34\hfill \end{array}$

## Solution

We will solve the first equation for $x$ and then substitute the expression into the second equation.

 Solve for x . Substitute into the other equation. Replace the x with 2 y − 2. Solve the resulting equation for y . Substitute y = 5 into x − 2 y = −2 to find x . The ordered pair is (8, 5). Check the ordered pair in both equations: $\begin{array}{cccc}\begin{array}{ccc}\hfill x-2y& =\hfill & -2\hfill \\ \hfill 8-2·5& \stackrel{?}{=}\hfill & -2\hfill \\ \hfill 8-10& \stackrel{?}{=}\hfill & -2\hfill \\ \hfill -2& =\hfill & -2\phantom{\rule{0.2em}{0ex}}✓\hfill \end{array}& & & \begin{array}{ccc}\hfill 3x+2y& =\hfill & 34\hfill \\ \hfill 3·8+2·5& \stackrel{?}{=}\hfill & 34\hfill \\ \hfill 24+10& \stackrel{?}{=}\hfill & 34\hfill \\ \hfill 34& =\hfill & 34\phantom{\rule{0.2em}{0ex}}✓\hfill \end{array}\end{array}$ The solution is (8, 5).

Solve the system by substitution. $\left\{\begin{array}{c}x-5y=13\hfill \\ 4x-3y=1\hfill \end{array}$

$\left(-2,-3\right)$

Solve the system by substitution. $\left\{\begin{array}{c}x-6y=-6\hfill \\ 2x-4y=4\hfill \end{array}$

$\left(6,2\right)$

When both equations are already solved for the same variable, it is easy to substitute!

Solve the system by substitution. $\left\{\begin{array}{c}y=-2x+5\hfill \\ y=\frac{1}{2}x\hfill \end{array}$

## Solution

Since both equations are solved for y , we can substitute one into the other.

 Substitute $\frac{1}{2}x$ for y in the first equation. Replace the y with $\frac{1}{2}x.$ Solve the resulting equation. Start by clearing the fraction. Solve for x . Substitute x = 2 into y = $\frac{1}{2}x$ to find y . The ordered pair is (2,1). Check the ordered pair in both equations: $\begin{array}{cccc}\begin{array}{ccc}\hfill y& =\hfill & \frac{1}{2}x\hfill \\ \hfill 1& \stackrel{?}{=}\hfill & \frac{1}{2}·2\hfill \\ \hfill 1& =\hfill & 1\phantom{\rule{0.2em}{0ex}}✓\hfill \end{array}& & & \begin{array}{ccc}\hfill y& =\hfill & -2x+5\hfill \\ \hfill 1& \stackrel{?}{=}\hfill & -2·2+5\hfill \\ \hfill 1& =\hfill & -4+5\hfill \\ \hfill 1& =\hfill & 1\phantom{\rule{0.2em}{0ex}}✓\hfill \end{array}\end{array}$ The solution is (2,1).

Solve the system by substitution. $\left\{\begin{array}{c}y=3x-16\hfill \\ y=\frac{1}{3}x\hfill \end{array}$

$\left(6,2\right)$

Solve the system by substitution. $\left\{\begin{array}{c}y=\text{−}x+10\hfill \\ y=\frac{1}{4}x\hfill \end{array}$

$\left(8,2\right)$

Be very careful with the signs in the next example.

Solve the system by substitution. $\left\{\begin{array}{c}4x+2y=4\hfill \\ 6x-y=8\hfill \end{array}$

## Solution

We need to solve one equation for one variable. We will solve the first equation for y .

 Solve the first equation for y . Substitute −2 x + 2 for y in the second equation. Replace the y with −2 x + 2. Solve the equation for x . Substitute $x=\frac{5}{4}$ into 4 x + 2 y = 4 to find y . The ordered pair is $\left(\frac{5}{4},-\frac{1}{2}\right).$ Check the ordered pair in both equations. $\begin{array}{cccc}\begin{array}{ccc}\hfill 4x+2y& =\hfill & 4\hfill \\ \hfill 4\left(\frac{5}{4}\right)+2\left(-\frac{1}{2}\right)& \stackrel{?}{=}\hfill & 4\hfill \\ \hfill 5-1& \stackrel{?}{=}\hfill & 4\hfill \\ \hfill 4& =\hfill & 4\phantom{\rule{0.2em}{0ex}}✓\hfill \\ \\ \\ \\ \\ \end{array}\hfill & & & \begin{array}{ccc}\hfill 6x-y& =\hfill & 8\hfill \\ \hfill 6\left(\frac{5}{4}\right)-\left(-\frac{1}{2}\right)& \stackrel{?}{=}\hfill & 8\hfill \\ \hfill \frac{15}{4}-\left(-\frac{1}{2}\right)& \stackrel{?}{=}\hfill & 8\hfill \\ \hfill \frac{16}{2}& \stackrel{?}{=}\hfill & 8\hfill \\ \hfill 8& =\hfill & 8\phantom{\rule{0.2em}{0ex}}✓\hfill \end{array}\hfill \end{array}$ The solution is $\left(\frac{5}{4},-\frac{1}{2}\right).$

Solve the system by substitution. $\left\{\begin{array}{c}x-4y=-4\hfill \\ -3x+4y=0\hfill \end{array}$

$\left(2,\frac{3}{2}\right)$

Solve the system by substitution. $\left\{\begin{array}{c}4x-y=0\hfill \\ 2x-3y=5\hfill \end{array}$

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

In [link] , it will take a little more work to solve one equation for x or y .

Solve the system by substitution. $\left\{\begin{array}{c}4x-3y=6\hfill \\ 15y-20x=-30\hfill \end{array}$

## Solution

We need to solve one equation for one variable. We will solve the first equation for x .

 Solve the first equation for x . Substitute $\frac{3}{4}y+\frac{3}{2}$ for x in the second equation. Replace the x with $\frac{3}{4}y+\frac{3}{2}.$ Solve for y .

Since 0 = 0 is a true statement, the system is consistent. The equations are dependent. The graphs of these two equations would give the same line. The system has infinitely many solutions.

Larry and Tom were standing next to each other in the backyard when Tom challenged Larry to guess how tall he was. Larry knew his own height is 6.5 feet and when they measured their shadows, Larry’s shadow was 8 feet and Tom’s was 7.75 feet long. What is Tom’s height?
6.25
Ciid
Wayne is hanging a string of lights 57 feet long around the three sides of his patio, which is adjacent to his house. the length of his patio, the side along the house, is 5 feet longer than twice it's width. Find the length and width of the patio.
Ciid
tyler
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Amara has to sell 120 televisions to make 29,000 of the salary of company B. 120 * TV 100= commission 12,000+ 17,000 = of company a salary 29,000 17000+
Ciid
Amara has to sell 120 televisions to make 29,000 of the salary of company B. 120 * TV 100= commission 12,000+ 17,000 = of company a salary 29,000
Ciid
I'm mathematics teacher from highly recognized university.
here a question professor How many soldiers are there in a group of 27 sailors and soldiers if there are four fifths as many sailors as soldiers? can you write out the college you went to with the name of the school you teach at and let me know the answer I've got it to be honest with you
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is anyone else having issues with the links not doing anything?
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chapter 1 foundations 1.2 exercises variables and algebraic symbols
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Joseph would like to make 12 pounds of a coffee blend at a cost of $6.25 per pound. He blends Ground Chicory at$4.40 a pound with Jamaican Blue Mountain at $8.84 per pound. How much of each type of coffee should he use? Samer 4x6.25=$25 coffee blend 4×4.40= $17.60 ground chicory 4x8.84= 35.36 blue mountain. In total they will spend for 12 pounds$77.96 they will spend in total
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How many soldiers are there in a group of 27 sailors and soldiers if there are four fifths many sailors as soldiers?
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June needs 45 gallons of punch for a party and has 2 different coolers to carry it in. The bigger cooler is 5 times as large as the smaller cooler. How many gallons can each cooler hold?