Let's solve this system of linear equations. 5x + 3y = 9 (1); x + 4y = 5 (2) Step 1: Multiply equation (2) by 5 to make the coefficients of x the same. 5(x + 4y) = 5(5); 5x + 20y = 25 (3) Step 2: Subtract equation (1) from equation (3) to eliminate x. (5x + 20y) − (5x + 3y) = 25 − 9; 5x + 20y − 5x − 3y = 16; 17y = 16 Step 3: Solve for y. y = 16/17 Step 4: Substitute the value of y into equation (2) to solve for x. x + 4(16/17) = 5; x + 64/17 = 5 Step 5: Isolate x. x = 5 − 64/17 To subtract, find a common denominator: x = (5 × 17)/(17) − 64/17; x = 85/17 − 64/17; x = 21/17 The solution to the system of equations is: x = 21/17, y = 16/17

5x+3y=9 and x+4y=5

June 1, 2026Mathematics

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Answer

x = \frac{21}{17}, y = \frac{16}{17}

Step-by-step solution

5 steps

Let's solve this system of linear equations.

5x + 3y = 9 \quad (1)

x + 4y = 5 \quad (2)

Multiply equation (2) by 5 to make the coefficients of $x$ the same.
$5(x + 4y) = 5(5)$ $5x + 20y = 25 \quad (3)$
Subtract equation (1) from equation (3) to eliminate $x$ .
$(5x + 20y) - (5x + 3y) = 25 - 9$ $5x + 20y - 5x - 3y = 16$ $17y = 16$
Solve for $y$ .
$y = \frac{16}{17}$
Substitute the value of $y$ into equation (2) to solve for $x$ .
$x + 4\left(\frac{16}{17}\right) = 5$ $x + \frac{64}{17} = 5$
Isolate $x$ .
$x = 5 - \frac{64}{17}$
To subtract, find a common denominator:
$x = \frac{5 \times 17}{17} - \frac{64}{17}$ $x = \frac{85}{17} - \frac{64}{17}$ $x = \frac{21}{17}$
The solution to the system of equations is:
$x = \frac{21}{17}, y = \frac{16}{17}$

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5x+3y=9 and x+4y=5

Answer

Step-by-step solution

Multiply equation (2) by 5 to make the coefficients of xxx the same.

Subtract equation (1) from equation (3) to eliminate xxx.

Solve for yyy.

Substitute the value of yyy into equation (2) to solve for xxx.

Isolate xxx.

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More Mathematics Questions

Multiply equation (2) by 5 to make the coefficients of $x$ the same.

Subtract equation (1) from equation (3) to eliminate $x$ .

Solve for $y$ .

Substitute the value of $y$ into equation (2) to solve for $x$ .

Isolate $x$ .