# 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.

Aziza is solving this equation-2(1+x)=4x+10
No. 3^32 -1 has exactly two divisors greater than 75 and less than 85 what is their product?
x^2+7x-19=0 has Two solutions A and B give your answer to 3 decimal places
3. When Jenna spent 10 minutes on the elliptical trainer and then did circuit training for20 minutes, her fitness app says she burned 278 calories. When she spent 20 minutes onthe elliptical trainer and 30 minutes circuit training she burned 473 calories. How manycalories does she burn for each minute on the elliptical trainer? How many calories doesshe burn for each minute of circuit training?
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Angelita
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p-2/3=5/6 how do I solve it with explanation pls
P=3/2
Vanarith
1/2p2-2/3p=5p/6
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4.5
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is y=7/5 a solution of 5y+3=10y-4
yes
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Nga and Lauren bought a chest at a flea market for $50. They re-finished it and then added a 350 % mark - up Makaila Reply$1750
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the sum of two Numbers is 19 and their difference is 15
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interesting
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4,2
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hola saben como aser un valor de la expresión
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integer greater than 2 and less than 12
2 < x < 12
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