Linear Equations with Two Variables – Examples and Practice

Step 1: Substitute: We have that $latex y=5$, therefore:

$latex 3x-4y=10$

$latex 3x-4(5)=10$

$latex 3x-20=10$

Step 2: Simplify: We have nothing to simplify:

Step 3: Solve for the variable: We add 20 to both sides:

$latex 3x-20=10$

$latex 3x-20+20=10+20$

$latex 3x=30$

Step 4: Solve: We divide both sides by 3:

$$\frac{3x}{3}=\frac{30}{3}$$

$latex x=10$

Step 1: Substitute: We have that $latex y=-3$, therefore:

$latex -3x+5y=-6$

$latex -3x+5(-3)=-6$

$latex -3x-15=-6$

Step 2: Simplify: We do not have like terms.

Step 3: Solve for the variable: We add 15 to both sides:

$latex -3x-15+15=-6+15$

$latex -3x=9$

Step 4: Solve: We divide both sides by -3:

$$\frac{-3x}{-3}=\frac{9}{-3}$$

$latex x=-3$

1. Substitute the value of the variable into the equation:

$latex 2x+4(2)=10-x+7$

$latex 2x+8=10-x+7$

2. Simplify: 

• We don’t have parentheses.
• We don’t have fractions.
• We combine like terms: $latex 2x+8=17-x$.

3. Isolate the variable: we move the 8 to the right and the –x to the left:

$latex 2x+8-8=17-x-8$

$latex 2x=9-x$

$latex 2x+x=9-x+x$

$latex 3x=9$

4. Solve for $latex x$: we divide both sides by 3: 

$$\frac{3}{3}x=\frac{{9}}{3}$$

$$x=\frac{{9}}{3}=3$$

5. Check the answer: substitute the values of the variables in the original equation:

$latex 2(3)+4(2)=10-(3)+7$

$latex 6+8=10-3+7$

$latex 14=14$

This is true.

Answer: $latex y=2,   x=3$.

1. Substitute the value of the variable: 

$$5(4)-2y+5=12-3(4)+15$$

$latex 20-2y+5=12-12+15$

2. Simplify: 

• We don’t have parentheses.
• We don’t have fractions.
• We combine like terms: $latex 25-2y=15$.

3. Isolate the variable: we move the 25 to the right: 

$latex 25-2y-25=15-25$

$latex -2y=-10$

4. Solve for $latex y$ completely: we divide both sides by-2: 

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

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

5. Check the answer: substitute the values of the variables in the original equation:

$$5(4)-2(5)+5=12-3(4)+15$$

$latex 20-10+5=12-12+15$

$latex 15=15$

This is true.

Answer: $latex y=5,   x=4$.

Step 1: Substitute: We have that $latex x=-2$, therefore:

$latex 4y+2(2y+3)=3(-2)-4$

$latex 4y+2(2y+3)=-6-4$

Step 2: Simplify: We expand the parentheses and combine like terms:

$latex 4y+2(2y+3)=-6-4$

$latex 4y+4y+6=-10$

$latex 8y+6=-10$

Step 3: Solve for the variable: We subtract 6 from both sides:

$latex 8y+6-6=-10-6$

$latex 8y=-16$

Step 4: Solve: We divide both sides by 8:

$$\frac{8y}{8}=\frac{16}{-8}$$

$latex y=-2$

We have no given value of any variable, so this equation has an infinite number of solutions. For example, suppose $latex x=1$, then we would have:

$latex 3x-2y=20+x$

$latex 3(1)-2y=20+1$

$latex 3-2y=21$

$latex -2y=21-3$

$latex -2y=18$

$latex y=-9$

Now, suppose that $latex x=2$, then we would have:

$latex 3x-2y=20+x$

$latex 3(2)-2y=20+2$

$latex 6-2y=22$

$latex -2y=22-6$

$latex -2y=16$

$latex y=-8$

We could continue with different values of $latex x$ and we would always get different answers. This means that there are an infinite number of solutions.

1. Substitute the value of the variable: 

 $latex 6+4(y+2)=10+2y+2(6)$

$latex 6+4(y+2)=10+2y+12$

2. Simplify: 

• Simplify parentheses: $latex 6+4y+8=10+2y+12$.
• We don’t have fractions.
• Combine like terms: $latex 4y+14=2y+22$.

3. Isolate the variable: we move the 14 to the right and the 2y to the left: 

$latex 4y+14-14=2y+22-14$

$latex 4y=2y+8$

$latex 4y-2y=2y+8-2y$

$latex 2y=8$

4. Solve for $latex y$: we divide both sides by 2: 

$$\frac{2}{2}y=\frac{{8}}{2}$$

$$y=\frac{{8}}{2}=4$$

5. Verify the answer: we substitute the values of the variables in the original equation:

$$6+4(4+2)=10+2(4)+2(6)$$

$latex 6+4(6)=10+8+12$

$latex 30=30$

This is true

Answer: $latex y=4,   x=6$

Step 1: Substitute: We substitute $latex z=5$ in the equation:

$$4y+2(5)=2(3y+10)+5-11$$

$$4y+10=2(3y+10)+5-11$$

Step 2: Simplify: We expand the parentheses and combine like terms:

$latex 4y+10=6y+20+5-11$

$latex 4y+10=6y+14$

Step 3: Solve for the variable: We subtract 10 and 6 y from both sides:

$latex 4y+10-10=6y+14-10$

$latex 4y=6y+4$

$latex 4y-6y=6y+4-6y$

$latex -2y=4$

Step 4: Solve: We divide both sides by -2:

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

$latex y=-2$

1. Substitute the value of the variable: 

$$\frac{1}{3}x+2(2)+\frac{1}{2}x=2+x+1$$

$$\frac{1}{3}x+4+\frac{1}{2}x=2+x+1$$

2. Simplify: 

• We don’t have parentheses.
• Simplify fractions: multiply by 6: $latex 2x+24+3x=12+6x+6$.
• Combine like terms: $latex 5x+24=18+6x$.

3. Isolate the variable: we move the 24 to the right and the 6x to the left: 

$latex 5x+24-24=18+6x-24$

$latex 5x=6x-6$

$latex 5x-6x=6x-6-6x$

$latex -x=-6$

4. Solve for $latex x$: we divide both sides by-1: 

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

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

5. Check the answer: substitute the values of the variables in the original equation:

$$\frac{1}{3}(6)+2(2)+\frac{1}{2}(6)=(2)+(6)+1$$

$latex 2+4+3=2+6+1$

$latex 9=9$

This is true.

Answer: $latex y=2,   x=6$

Step 1: Substitute: We have that $latex y=-3$, therefore:

$$\frac{-3+1}{2}+2x=2(2(-3)+6)+x+2$$

$$\frac{-3+1}{2}+2x=2(-6+6)+x+2$$

Step 2: Simplify: We combine like terms and simplify:

$$\frac{-2}{2}+2x=2(0)+x+2$$

$latex -1+2x=x+2$

Step 3: Solve for the variable: We add 1 and subtract x from both sides:

$latex -1+2x+1=x+2+1$

$latex 2x=x+3$

$latex 2x-x=x+3-x$

$latex x=3$

Step 4: Solve: We have already found the solution:

$latex x=3$

Step 1: Substitute: In this case, we do not have any given value, so automatically, the equation has an infinite number of solutions. For example, suppose we have $latex x = 0$, then we would have:

$latex 2x+2y=3x+10$

$latex 2(0)+2y=3(0)+10$

$latex 2y=10$

$latex y=5$

If we now have $latex x = 1$, we have:

$latex 2x+2y=3x+10$

$latex 2(1)+2y=3(1)+10$

$latex 2+2y=3+10$

$latex 2+2y=13$

$latex 2y=15$

$latex y=15/2$

We could continue with different values and each time we would get different results, so by not having a specified value of a variable, the equation has infinite solutions.