# 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.

the proudact of 3x^3-5×^2+3 and 2x^2+5x-4 in z7[x]/ is
?
Choli
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44.19%
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p mulripied-5 and add 30 to it
Tausif
Tausif
Can you explain further
p mulripied-5 and add to 30
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TAKE OFF THE LAST DIGIT AND MULTIPLY IT 9. SUBTRACT IT THE DIGITS YOU HAVE LEFT. IF THE ANSWER DIVIDES BY 13(OR IS ZERO), THEN YOUR ORIGINAL NUMBER WILL ALSO DIVIDE BY 13!IS DIVISIBLE BY 13
BAINAMA
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snigdha
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Esther
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Orlando
hello, I have algebra phobia. Subtracting negative numbers always seem to get me confused.
what do you need help in?
Felix
Heather
look at the numbers if they have different signs, it's like subtracting....but you keep the sign of the largest number...
Felix
for example.... -19 + 7.... different signs...subtract.... 12 keep the sign of the "largest" number 19 is bigger than 7.... 19 has the negative sign... Therefore, -12 is your answer...
Felix
—12
Thanks Felix.l also get confused with signs.
Esther
Thank you for this
Shatey
ty
Graham
think about it like you lost $19 (-19), then found$7(+7). Totally you lost just $12 (-12) Annushka I used to struggle a lot with negative numbers and math in general what I typically do is look at it in terms of money I have -$5 in my account I then take out 5 more dollars how much do I have in my account well-\$10 ... I also for a long time would draw it out on a number line to visualize it
Meg
practicing with smaller numbers to understand then working with larger numbers helps too and the song/rhyme same sign add and keep opposite signs subtract keep the sign of the bigger # then you'll be exact
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Bruce drives his car for his job. The equation R=0.575m+42 models the relation between the amount in dollars, R, that he is reimbursed and the number of miles, m, he drives in one day. Find the amount Bruce is reimbursed on a day when he drives 220 miles
168.50=R
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john is 5years older than wanjiru.the sum of their years is27years.what is the age of each
46
mustee
j 17 w 11
Joseph
john is 16. wanjiru is 11.
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27-5=22 22÷2=11 11+5=16
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I don't see where the answers are.
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3
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